Class 9th Mathematics AP Board Solution
Exercise 8.1- State whether the statements are True or False.(i) Every parallelogram is a…
- Complete the following table by writing (YES) if the property holds for the…
- ABCD is trapezium in which AB || CD. If AD = BC, show that ∠A = ∠B and ∠C = ∠D.…
- The four angles of a quadrilateral are in the ratio 1: 2:3:4. Find the measure…
- ABCD is a rectangle AC is diagonal. Find the angles of ∆ACD. Give reasons.…
Exercise 8.2- In the adjacent figure ABCD is a parallelogram ABEF is a rectangle show that…
- Show that the diagonals of a rhombus divide it into four congruent triangles.…
- In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O.Prove that…
Exercise 8.3- The opposite angles of a parallelogram are (3x-2)^{0} and (x+48)^{0}…
- Find the measure of all the angles of a parallelogram, if one angle is 24° less…
- In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side…
- In the adjacent figure ABCD is a parallelogram P, Q are the midpoints of sides…
- ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and…
- ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C…
- In delta^{3}abc and DEF, AB || DE, AB=DE; BC = EF and BC || EF. Vertices A, B…
- ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are…
- ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA…
Exercise 8.4- ABC is a triangle. D is a point on AB such that ad = {1}/{4} ab and E is a…
- ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA…
- Show that the figure formed by joining the midpoints of sides of a rhombus…
- In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC…
- Show that the line segments joining the midpoints of the opposite sides of a…
- ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse…
- State whether the statements are True or False.(i) Every parallelogram is a…
- Complete the following table by writing (YES) if the property holds for the…
- ABCD is trapezium in which AB || CD. If AD = BC, show that ∠A = ∠B and ∠C = ∠D.…
- The four angles of a quadrilateral are in the ratio 1: 2:3:4. Find the measure…
- ABCD is a rectangle AC is diagonal. Find the angles of ∆ACD. Give reasons.…
- In the adjacent figure ABCD is a parallelogram ABEF is a rectangle show that…
- Show that the diagonals of a rhombus divide it into four congruent triangles.…
- In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O.Prove that…
- The opposite angles of a parallelogram are (3x-2)^{0} and (x+48)^{0}…
- Find the measure of all the angles of a parallelogram, if one angle is 24° less…
- In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side…
- In the adjacent figure ABCD is a parallelogram P, Q are the midpoints of sides…
- ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and…
- ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C…
- In delta^{3}abc and DEF, AB || DE, AB=DE; BC = EF and BC || EF. Vertices A, B…
- ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are…
- ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA…
- ABC is a triangle. D is a point on AB such that ad = {1}/{4} ab and E is a…
- ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA…
- Show that the figure formed by joining the midpoints of sides of a rhombus…
- In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC…
- Show that the line segments joining the midpoints of the opposite sides of a…
- ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse…
Exercise 8.1
Question 1.State whether the statements are True or False.
(i) Every parallelogram is a trapezium ( )
(ii) All parallelograms are quadrilaterals ( )
(iii) All trapeziums are parallelograms ( )
(iv) A square is a rhombus ( )
(v) Every rhombus is a square ( )
(vi) All parallelograms are rectangles ( )
Answer:(i) [True]
⇒ The trapezium has a property
One pair of opposite sides are parallel
This is conveyed by parallelogram
as they have 2 pair of parallel sides
∴ Every parallelogram is a trapezium
(ii) [True]
⇒ The quadrilateral is known as
Polygon having four sides
This is conveyed by parallelogram
as they are also four sided polygon.
∴ Every parallelogram is a Quadrilateral
(iii) [False]
⇒ The parallelogram has a property
Both pair of opposite sides are parallel and equal
Which is not conveyed by trapezium
as they have only 1 pair of parallel sides
∴ Every trapezium is not parallelogram
(iv) [True]
⇒ The Rhombus has a property
Diagonal of rhombus bisect each other at 90°
This is conveyed by square
as their diagonal also bisect each other at 90°
∴ A square is a rhombus
(v) [False]
⇒ The square has a property
All angles of square are equal and 90°
And diagonals are equal and perpendicular bisector to each
other
Which is not conveyed by Rhombus
as their diagonal are perpendicular bisector but not equal
And their angles are also not equal to 90°
∴ Every rhombus is not a square.
(vi) [False]
⇒ The rectangle has a property
Every angle of rectangle is 90°
Which is not conveyed by parallelogram
as it have only opposite side angles are equal not 90°
∴ Every parallelogram are not rectangle.
Question 2.Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
Answer:
Question 3.ABCD is trapezium in which AB || CD. If AD = BC, show that ∠A = ∠B and ∠C = ∠D.
Answer:
Given:- ABCD is a isosceles trapezium
Trapezium with one pair of sides is parallel
And other pair of sides is equal[AD=BC].
Formula used:- SSS congruency property
If all 3 sides of triangle are equal to all 3 sides of
other triangle
Then; Both triangles are congruent
Solution:- In trapezium ABCD
If AB||CD
And; AD=BC
∴ ABCD is a isosceles trapezium
⇒ If ABCD is a isosceles trapezium
Then;
Diagonals of ABCD must be equal
∴ AC=BD
In Δ ABD and Δ ABC
AC=BD [∵ ABCD is isosceles trapezium]
AD=BC [Given]
AB=AB [Common line in both triangle]
∴ Both triangles are congruent by SSS property
∆ABD ≅ ∆ABC
⇒ If both triangles are congruent
Then there were also be equal.
∴ ∠ A=∠ B
As ABCD is trapezium and AB||CD
∠ A+∠ D=180° and ∠ C+∠ B=180°
∠ A =180° - ∠ D and ∠ B=180° - ∠ C
If ∠ A=∠ B;
180° - ∠ D=180° - ∠ C
180° +∠ C - 180° = ∠ D
∴ ∠ C=∠ D;
Conclusion:- If ABCD is trapezium in which AD = BC, then ∠A = ∠B and ∠C = ∠D.
Question 4.The four angles of a quadrilateral are in the ratio 1: 2:3:4. Find the measure of each angle of the quadrilateral.
Answer:
Given . Angles of a quadrilateral are in the ratio 1: 2:3:4
Solution .
Let any quadrilateral be ABCD
Then
Sum of all angles of quadrilateral is 360°
∴ ∠ A + ∠ B + ∠ C + ∠ D=360°
If angle A,B,C,D are in ratio 1:2:3:4
Then,
1x + 2x + 3x + 4x =360°
10x=360°
x= 
x= 36°
all angles are 1x,2x,3x,4x
Conclusion.
∴ Angles are 36° , 72° , 108° , 144°
Question 5.ABCD is a rectangle AC is diagonal. Find the angles of ∆ACD. Give reasons.
Answer:
Given:- ABCD is a rectangle
Formula used:- All angles of rectangle are 90°
In right angle triangle
Pythagoras theorem a2+b2=c2
Sin(θ)=
Solution:-
In ΔACD
∠ D=90° [All angles of rectangle are 90° ]
∠ ACD=θ1
And; Sin(θ1) = 
=
θ1=arc sin()
∠ DAC=θ2
And; Sin(θ2) = =
θ1=arc sin(
)
Conclusion:-
The angles of Δ ACD are :-
∠ D=90°
∠ DAC=arc sin()
∠ ACD=arc sin(
)
State whether the statements are True or False.
(i) Every parallelogram is a trapezium ( )
(ii) All parallelograms are quadrilaterals ( )
(iii) All trapeziums are parallelograms ( )
(iv) A square is a rhombus ( )
(v) Every rhombus is a square ( )
(vi) All parallelograms are rectangles ( )
Answer:
(i) [True]
⇒ The trapezium has a property
One pair of opposite sides are parallel
This is conveyed by parallelogram
as they have 2 pair of parallel sides
∴ Every parallelogram is a trapezium
(ii) [True]
⇒ The quadrilateral is known as
Polygon having four sides
This is conveyed by parallelogram
as they are also four sided polygon.
∴ Every parallelogram is a Quadrilateral
(iii) [False]
⇒ The parallelogram has a property
Both pair of opposite sides are parallel and equal
Which is not conveyed by trapezium
as they have only 1 pair of parallel sides
∴ Every trapezium is not parallelogram
(iv) [True]
⇒ The Rhombus has a property
Diagonal of rhombus bisect each other at 90°
This is conveyed by square
as their diagonal also bisect each other at 90°
∴ A square is a rhombus
(v) [False]
⇒ The square has a property
All angles of square are equal and 90°
And diagonals are equal and perpendicular bisector to each
other
Which is not conveyed by Rhombus
as their diagonal are perpendicular bisector but not equal
And their angles are also not equal to 90°
∴ Every rhombus is not a square.
(vi) [False]
⇒ The rectangle has a property
Every angle of rectangle is 90°
Which is not conveyed by parallelogram
as it have only opposite side angles are equal not 90°
∴ Every parallelogram are not rectangle.
Question 2.
Complete the following table by writing (YES) if the property holds for the particular Quadrilateral and (NO) if property does not holds.
Answer:
Question 3.
ABCD is trapezium in which AB || CD. If AD = BC, show that ∠A = ∠B and ∠C = ∠D.
Answer:
Given:- ABCD is a isosceles trapezium
Trapezium with one pair of sides is parallel
And other pair of sides is equal[AD=BC].
Formula used:- SSS congruency property
If all 3 sides of triangle are equal to all 3 sides of
other triangle
Then; Both triangles are congruent
Solution:- In trapezium ABCD
If AB||CD
And; AD=BC
∴ ABCD is a isosceles trapezium
⇒ If ABCD is a isosceles trapezium
Then;
Diagonals of ABCD must be equal
∴ AC=BD
In Δ ABD and Δ ABC
AC=BD [∵ ABCD is isosceles trapezium]
AD=BC [Given]
AB=AB [Common line in both triangle]
∴ Both triangles are congruent by SSS property
∆ABD ≅ ∆ABC
⇒ If both triangles are congruent
Then there were also be equal.
∴ ∠ A=∠ B
As ABCD is trapezium and AB||CD
∠ A+∠ D=180° and ∠ C+∠ B=180°
∠ A =180° - ∠ D and ∠ B=180° - ∠ C
If ∠ A=∠ B;
180° - ∠ D=180° - ∠ C
180° +∠ C - 180° = ∠ D
∴ ∠ C=∠ D;
Conclusion:- If ABCD is trapezium in which AD = BC, then ∠A = ∠B and ∠C = ∠D.
Question 4.
The four angles of a quadrilateral are in the ratio 1: 2:3:4. Find the measure of each angle of the quadrilateral.
Answer:
Given . Angles of a quadrilateral are in the ratio 1: 2:3:4
Solution .
Let any quadrilateral be ABCD
Then
Sum of all angles of quadrilateral is 360°
∴ ∠ A + ∠ B + ∠ C + ∠ D=360°
If angle A,B,C,D are in ratio 1:2:3:4
Then,
1x + 2x + 3x + 4x =360°
10x=360°
x=
x= 36°
all angles are 1x,2x,3x,4x
Conclusion.
∴ Angles are 36° , 72° , 108° , 144°
Question 5.
ABCD is a rectangle AC is diagonal. Find the angles of ∆ACD. Give reasons.
Answer:
Given:- ABCD is a rectangle
Formula used:- All angles of rectangle are 90°
In right angle triangle
Pythagoras theorem a2+b2=c2
Sin(θ)=
Solution:-
In ΔACD
∠ D=90° [All angles of rectangle are 90° ]
∠ ACD=θ1
And; Sin(θ1) = =
θ1=arc sin()
∠ DAC=θ2
And; Sin(θ2) = =
θ1=arc sin()
Conclusion:-
The angles of Δ ACD are :-
∠ D=90°
∠ DAC=arc sin()
∠ ACD=arc sin()
Exercise 8.2
Question 1.In the adjacent figure ABCD is a parallelogram ABEF is a rectangle show that ∆AFD ≅ ∆BEC.
Answer:Given :- ABCD is a parallelogram and ABEF is a rectangle
Formula used :-
SAS congruency rule
If two sides of triangle and angle made by the 2 sides are equal then both the triangles are congruent
Solution :-
In ∆AFD and ∆BEC
*AF=BE [opposite sides of rectangle are equal]
*AD=BC [opposite sides of parallelogram are equal]
As angle of rectangle is 90°
∠ DAB +∠ FAD=90°
∠ DAB=90° - ∠ FAD -----1
Sum of corresponding angles of parallelogram is 180°
∠ DAB+∠ ABC =180°
∠ DAB+∠ ABE+∠ EBC=180°
90° - ∠ FAD + 90° +∠ EBC=180° ∵ putting value from 1
180° +∠ EBC-∠ FAD=180°
∠ EBC - ∠ FAD=0
∠ EBC = ∠ FAD
⇒ Hence;
By SAS property both triangles are congruent
Conclusion :-
∆AFD ≅ ∆BEC
Question 2.Show that the diagonals of a rhombus divide it into four congruent triangles.
Answer:
Given :- ABCD is a rhombus
Formula used :-
*SSS congruency rule
If all sides of both triangles are equal then both triangles are congruent
*properties of rhombus
Solution :-
In Δ AOD and Δ COB
AO=OC [diagonal of Rhombus bisect each other]
OD=OB [diagonal of Rhombus bisect each other]
AD=BC [all sides of rhombus are equal]
∴ Δ AOD ≅ Δ COB
In Δ AOB and Δ COD
AO=OC [diagonal of Rhombus bisect each other]
OD=OB [diagonal of Rhombus bisect each other]
CD=BA [all sides of rhombus are equal]
∴ Δ AOB ≅ Δ COD
In Δ AOB and Δ AOD
AO=AO [Common in both triangles]
OD=OB [diagonal of Rhombus bisect each other]
AD=AB [all sides of rhombus are equal]
∴ Δ AOB ≅ Δ AOD
∴ All four triangles divide by diagonals of triangle are congruent
Question 3.In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O.
Prove that 
Answer:
Given :- ABCD is a quadrilateral
∠ OCB= ∠ OCD=∠C/2 [OC is bisector of ∠ C]
∠ ODA= ∠ ODC=∠D/2 [OD is bisector of ∠ D]
Formula Used:- ∠ A+ ∠ B+∠ C+ ∠ D=360°
Solution :-
In Δ COD
∠ OCD+ ∠ COD +∠ ODC=180°
∠D/2 + ∠C/2 +∠ COD = 180°
(∠ D+∠ C)/2 + ∠ COD = 180°
If;
∠ A+ ∠ B+∠ C+ ∠ D=360°
∠ C+ ∠ D=360° -(∠ A+ ∠ B)
+ ∠ COD=180°
180° -
+ ∠ COD=180°
∠ COD=
In the adjacent figure ABCD is a parallelogram ABEF is a rectangle show that ∆AFD ≅ ∆BEC.
Answer:
Given :- ABCD is a parallelogram and ABEF is a rectangle
Formula used :-
SAS congruency rule
If two sides of triangle and angle made by the 2 sides are equal then both the triangles are congruent
Solution :-
In ∆AFD and ∆BEC
*AF=BE [opposite sides of rectangle are equal]
*AD=BC [opposite sides of parallelogram are equal]
As angle of rectangle is 90°
∠ DAB +∠ FAD=90°
∠ DAB=90° - ∠ FAD -----1
Sum of corresponding angles of parallelogram is 180°
∠ DAB+∠ ABC =180°
∠ DAB+∠ ABE+∠ EBC=180°
90° - ∠ FAD + 90° +∠ EBC=180° ∵ putting value from 1
180° +∠ EBC-∠ FAD=180°
∠ EBC - ∠ FAD=0
∠ EBC = ∠ FAD
⇒ Hence;
By SAS property both triangles are congruent
Conclusion :-
∆AFD ≅ ∆BEC
Question 2.
Show that the diagonals of a rhombus divide it into four congruent triangles.
Answer:
Given :- ABCD is a rhombus
Formula used :-
*SSS congruency rule
If all sides of both triangles are equal then both triangles are congruent
*properties of rhombus
Solution :-
In Δ AOD and Δ COB
AO=OC [diagonal of Rhombus bisect each other]
OD=OB [diagonal of Rhombus bisect each other]
AD=BC [all sides of rhombus are equal]
∴ Δ AOD ≅ Δ COB
In Δ AOB and Δ COD
AO=OC [diagonal of Rhombus bisect each other]
OD=OB [diagonal of Rhombus bisect each other]
CD=BA [all sides of rhombus are equal]
∴ Δ AOB ≅ Δ COD
In Δ AOB and Δ AOD
AO=AO [Common in both triangles]
OD=OB [diagonal of Rhombus bisect each other]
AD=AB [all sides of rhombus are equal]
∴ Δ AOB ≅ Δ AOD
∴ All four triangles divide by diagonals of triangle are congruent
Question 3.
In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O.
Prove that
Answer:
Given :- ABCD is a quadrilateral
∠ OCB= ∠ OCD=∠C/2 [OC is bisector of ∠ C]
∠ ODA= ∠ ODC=∠D/2 [OD is bisector of ∠ D]
Formula Used:- ∠ A+ ∠ B+∠ C+ ∠ D=360°
Solution :-
In Δ COD
∠ OCD+ ∠ COD +∠ ODC=180°
∠D/2 + ∠C/2 +∠ COD = 180°
(∠ D+∠ C)/2 + ∠ COD = 180°
If;
∠ A+ ∠ B+∠ C+ ∠ D=360°
∠ C+ ∠ D=360° -(∠ A+ ∠ B)
+ ∠ COD=180°
180° - + ∠ COD=180°
∠ COD=
Exercise 8.3
Question 1.The opposite angles of a parallelogram are 
and 
Find the measure of each angle of the parallelogram.
Answer:
Given:- Opposite angles of parallelogram and
Formula used:- opposite angles of a parallelogram are equal
Corresponding angles sum is 180°
Solution:-
Let ∠ A and ∠ C
∠ A = ∠ C [opposite angles of a parallelogram are equal]
3x-2=x+48
3x – x=48+2
2x=50
x=25
Then;
∠ A=∠ C=x+48=25+ 48 = 73°
∠ A+∠ B=180°
∠ B=180° - 73°
∠ B= 107°
∠ B = ∠ D [opposite angles of a parallelogram are equal]
Conclusion:-
Angles of parallelogram = 73°, 107°, 73°, 107°
Question 2.Find the measure of all the angles of a parallelogram, if one angle is 24° less than the twice of the smallest angle.
Answer:
Given:- one angle is 24° less than the twice of the smallest angle
Formula used:- opposite angles of a parallelogram are equal
Corresponding angles sum is 180°
Solution:-
Let ∠ A and ∠ C be smallest because
∠ A = ∠ C [opposite angles of a parallelogram are equal]
2∠ A+ 24° =∠ B
∠ A+∠ B=180° [Sum of corresponding angles is 180° ]
∠ B=180°-∠ A
Then,
2∠ A+ 24° =180° -∠ A
3∠ A =180° - 24°
3∠ A=156°
∠ A=
=52°
∠ B=180° -∠ A
∠ B= 180° -52°
∠ B=128°
∠ B = ∠ D [opposite angles of a parallelogram are equal]
Conclusion:-
∴ Angles of parallelogram = 52°, 128°, 52°, 128°
Question 3.In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB.
Answer:Given :- ABCD is a parallelogram
And CE=EB
Formula used:- ASA congruency property
If 2 angles and one side between them in two
triangles are equal then both triangle are congruent
Solutions :-
In Δ DCE and Δ FBE
∠ DEC=∠ BEF [Vertically opposite angles]
As DC || BF
∠ ECD=∠ EBF [Alternate angles]
And;
CE=EB [Given]
∴ Δ DCE ≅ Δ FBE
As Δ DCE ≅ Δ FBE
Then
DC=BF
And DC=AB [opposite sides of parallelogram are equal]
Hence ;
AB=BF
AF=AB+BF
AF=2AB.
Hence Proved;
Question 4.In the adjacent figure ABCD is a parallelogram P, Q are the midpoints of sides AB and DC respectively. Show that PBCQ is also a parallelogram.
Answer:Given:-ABCD is a parallelogram
P,Q are the midpoints of sides AB and DC respectively.
Solution:-
In parallelogram ABCD
If P,Q are the midpoints of sides AB and DC respectively
∴ DC=2QC=2QD
∴ AB=2AP=2PB
If DC||AB
Then;
QD||AP and QC||PB
Line PQ divides the parallelogram in 2 equal parts
Because P and Q are midpoints
∴ AD||PQ||CB
If;
PQ||CB and QC||PB
Then, PBCQ is also a parallelogram
Question 5.ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD || BA as shown in the figure. Show that
(i) ∠DAC = ∠BCA
(ii) ABCD is a parallelogram
Answer:Given:- AB=AC and ABC is isosceles triangle
∠ QAD=∠ DAC
CD||BA
Formula used:- sum of angles of triangle is 180°
Solution:-
As Δ ABC is a isosceles triangle
∠ ABC=∠ ACB
∠ ABC+∠ ACB+∠ CAB=180°
2∠ ACB+∠ CAB=180°
*As BAQ is a straight line
∠ CAB+∠ DAC+∠ QAD=180°
*∠ QAD=∠ DAC [Given]
∠ CAB=180° - 2∠ DAC
Putting value of ∠ CAB in above equation 1
2∠ ACB+180° - 2∠ DAC=180°
2∠ ACB =2∠ DAC
∠ ACB =∠ DAC
If;
There is ∠ ACB =∠ DAC and AC is the transverse
∴ these are equal by alternate angles
And AD||BC
In ABCD
If;
AD||BC & CD||BA
If both pair of sides of quadrilateral are parallel
Then the quadrilateral is parallelogram
Conclusion:-
ABCD is a parallelogram
And ∠DAC = ∠BCA
Question 6.ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that
(i) Δ APB ≅ Δ CQD
(ii) AP = CQ
Answer:Given:- ABCD is parallelogram
Formula used :- AAS property
If 2 angles and any one side is are equal in both triangle
Then both triangles are congruent
Solution:-
In Δ APB and Δ CQD
∠ CQD=∠ APB=90°
If ABCD is a parallelogram
∠ CDB=∠ DBA [alternate angles as DC||AB]
AB=DC [opposite sides of parallelogram are equal]
By AAS property
∴ Δ APB ≅ Δ CQD
If Δ APB ≅ Δ CQD
Then
AP=CQ
Conclusion:-
In parallelogram ABCD; Δ APB ≅ Δ CQD
Question 7.In 
and DEF, AB || DE, AB=DE; BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that
(i) ABED is a parallelogram
(ii) BCFE is a parallelogram
(iii) AC = DF
(iv) ∆ABC ≅ ∆DEF
Answer:Given:- AB||DE , AB=DE; BC||EF, BC=EF
Formula used:- SSS congruency rule
If all 3 sides of both triangle are equal then
Both triangles are congruent.
Solution:-
(1) In ABED
AB||DE
AB=DE
∴ ABED is a parallelogram
∵ If one pair of side in quadrilateral is equal and parallel
Then the quadrilateral is parallelogram.
(2) In BCFE
BC||EF
BC=EF
∴ BCEF is a parallelogram
∵ If one pair of side in quadrilateral is equal and parallel
Then the quadrilateral is parallelogram.
(3) As ABED is a parallelogram
BE||AD , BE=AD ;
As BCFE is a parallelogram
BE||CF , BE=CF ;
∴ By concluding both above statements
AD||CF and AD=CF
∴ ACFD is a parallelogram
∵ If one pair of side in quadrilateral is equal and parallel
Then the quadrilateral is parallelogram
If ACDF is a parallelogram ;
Then
AC=DF [opposite sides of parallelogram are equal ]
(4) In Δ ABC and Δ DEF
AB=DE [Given]
BC=EF [Given]
AC=DF [Proved above]
Δ ABC ≅ Δ DEF [SSS congruency rule]
Question 8.ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are the points of tri section of the diagonal BD. Prove that CQ || AP and also AC bisects PQ (see figure).
Answer:Given:- ABCD is parallelogram
BP=BD/3
DQ=BD/3
Formula used:- SAS congruency property
If 2 sides and angle between the two sides of both
the triangle are equal, then both triangle are congruent
Solution:-
⇒ In parallelogram ABCD
AO=OC and DO=OB [diagonal of parallelogram bisect each other]
As DO=OB
Where DO=DQ+OQ
OB=OP+PB
∴ DQ+OQ=OP+PB
⇒ 
+OQ=OP+
⇒ OQ=OP+ -
⇒ OQ=OP
PQ = OP+OQ = 2(OQ) = 2(OP)
∴ AC diagonal bisect PQ
⇒ In Δ QOC and Δ POA
OQ=PO [Proven above]
AO=OC [Diagonal of parallelogram bisect each other]
∠ QOC=∠ AOP [vertically opposite angles]
Hence Δ QOC ≅ Δ POA
∴ ∠ CQO=∠ OPA
If QC and PA are 2 lines
And QP is the transverse
And ∠ CQO=∠ OPA by Alternate angles
∴ QC||PA
Conclusion:-
CQ||PA and CA is bisector of PQ.
Question 9.ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.
Answer:Given:- *ABCD is a square
*E, F, G and H are the mid points of AB, BC, CD and DA
respectively
*AE = BF = CG = DH
Formula used:- *Isosceles Δ property
⇒ If 2 sides of triangle are equal then
Corresponding angle will also be equal
*Properties of quadrilateral to be square
⇒ All sides are equal
⇒ All angles are 90°
Solution:-
In Δ AHE, Δ EBF, Δ FCG, Δ DHG
⇒ AE = BF = CG = DH [Given]
If;
⇒ AE=EB [E is midpoint of AB]
⇒ BF=FC [F is midpoint of BC]
⇒ CG=GD [G is midpoint of CD]
⇒ DH=HA [H is midpoint of DA]
⇒ AE = BF = CG = DH
On replacing every part we get;
⇒ EB=FC=GD=HA;
⇒ ∠ A+∠ B+∠ C+∠ D=90° [All angles of square are 90°]
Hence;
All triangles Δ AHE, Δ EBF, Δ FCG, Δ DHG
are congruent by SAS property
∴ Δ AHE≅ Δ EBF≅ Δ FCG ≅ Δ DHG
⇒ HE=EF=FG=GH [All triangles are congruent]
In Δ AHE, Δ EBF, Δ FCG, Δ DHG
∵ all sides of square are equal and after the midpoint of each sides
Every half side of square are equal to half of other sides.
HA=AE , EB=FB ,FC=GC ,HD=DG
∴ All Δ AHE, Δ EBF, Δ FCG, Δ DHG are isosceles
⇒ as central angle of all triangle is 90°
It makes all Δ AHE, Δ EBF, Δ FCG, Δ DHG are right angle isosceles Δ
∴ all corresponding angles of equal side will be 45°
∠ AHE=∠ BEF=∠ CFG=∠ DHG=∠ AEH=∠ BFE=∠ CGF=∠ DGH=45°
⇒ as AB is straight line
Then; ∠ AEH+∠ HEF+∠ BEF=180°
45°+∠ HEF+45° =180°
∠ HEF=180° -90° =90°
Similarly ;
∠ EFG=90°
∠ FGH=90°
∠ GHE=90°
Conclusion:-
All angles are 90° and all sides are equal of quadrilateral
Hence quadrilateral is square
The opposite angles of a parallelogram are and Find the measure of each angle of the parallelogram.
Answer:
Given:- Opposite angles of parallelogram and
Formula used:- opposite angles of a parallelogram are equal
Corresponding angles sum is 180°
Solution:-
Let ∠ A and ∠ C
∠ A = ∠ C [opposite angles of a parallelogram are equal]
3x-2=x+48
3x – x=48+2
2x=50
x=25
Then;
∠ A=∠ C=x+48=25+ 48 = 73°
∠ A+∠ B=180°
∠ B=180° - 73°
∠ B= 107°
∠ B = ∠ D [opposite angles of a parallelogram are equal]
Conclusion:-
Angles of parallelogram = 73°, 107°, 73°, 107°
Question 2.
Find the measure of all the angles of a parallelogram, if one angle is 24° less than the twice of the smallest angle.
Answer:
Given:- one angle is 24° less than the twice of the smallest angle
Formula used:- opposite angles of a parallelogram are equal
Corresponding angles sum is 180°
Solution:-
Let ∠ A and ∠ C be smallest because
∠ A = ∠ C [opposite angles of a parallelogram are equal]
2∠ A+ 24° =∠ B
∠ A+∠ B=180° [Sum of corresponding angles is 180° ]
∠ B=180°-∠ A
Then,
2∠ A+ 24° =180° -∠ A
3∠ A =180° - 24°
3∠ A=156°
∠ A==52°
∠ B=180° -∠ A
∠ B= 180° -52°
∠ B=128°
∠ B = ∠ D [opposite angles of a parallelogram are equal]
Conclusion:-
∴ Angles of parallelogram = 52°, 128°, 52°, 128°
Question 3.
In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB.
Answer:
Given :- ABCD is a parallelogram
And CE=EB
Formula used:- ASA congruency property
If 2 angles and one side between them in two
triangles are equal then both triangle are congruent
Solutions :-
In Δ DCE and Δ FBE
∠ DEC=∠ BEF [Vertically opposite angles]
As DC || BF
∠ ECD=∠ EBF [Alternate angles]
And;
CE=EB [Given]
∴ Δ DCE ≅ Δ FBE
As Δ DCE ≅ Δ FBE
Then
DC=BF
And DC=AB [opposite sides of parallelogram are equal]
Hence ;
AB=BF
AF=AB+BF
AF=2AB.
Hence Proved;
Question 4.
In the adjacent figure ABCD is a parallelogram P, Q are the midpoints of sides AB and DC respectively. Show that PBCQ is also a parallelogram.
Answer:
Given:-ABCD is a parallelogram
P,Q are the midpoints of sides AB and DC respectively.
Solution:-
In parallelogram ABCD
If P,Q are the midpoints of sides AB and DC respectively
∴ DC=2QC=2QD
∴ AB=2AP=2PB
If DC||AB
Then;
QD||AP and QC||PB
Line PQ divides the parallelogram in 2 equal parts
Because P and Q are midpoints
∴ AD||PQ||CB
If;
PQ||CB and QC||PB
Then, PBCQ is also a parallelogram
Question 5.
ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD || BA as shown in the figure. Show that
(i) ∠DAC = ∠BCA
(ii) ABCD is a parallelogram
Answer:
Given:- AB=AC and ABC is isosceles triangle
∠ QAD=∠ DAC
CD||BA
Formula used:- sum of angles of triangle is 180°
Solution:-
As Δ ABC is a isosceles triangle
∠ ABC=∠ ACB
∠ ABC+∠ ACB+∠ CAB=180°
2∠ ACB+∠ CAB=180°
*As BAQ is a straight line
∠ CAB+∠ DAC+∠ QAD=180°
*∠ QAD=∠ DAC [Given]
∠ CAB=180° - 2∠ DAC
Putting value of ∠ CAB in above equation 1
2∠ ACB+180° - 2∠ DAC=180°
2∠ ACB =2∠ DAC
∠ ACB =∠ DAC
If;
There is ∠ ACB =∠ DAC and AC is the transverse
∴ these are equal by alternate angles
And AD||BC
In ABCD
If;
AD||BC & CD||BA
If both pair of sides of quadrilateral are parallel
Then the quadrilateral is parallelogram
Conclusion:-
ABCD is a parallelogram
And ∠DAC = ∠BCA
Question 6.
ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that
(i) Δ APB ≅ Δ CQD
(ii) AP = CQ
Answer:
Given:- ABCD is parallelogram
Formula used :- AAS property
If 2 angles and any one side is are equal in both triangle
Then both triangles are congruent
Solution:-
In Δ APB and Δ CQD
∠ CQD=∠ APB=90°
If ABCD is a parallelogram
∠ CDB=∠ DBA [alternate angles as DC||AB]
AB=DC [opposite sides of parallelogram are equal]
By AAS property
∴ Δ APB ≅ Δ CQD
If Δ APB ≅ Δ CQD
Then
AP=CQ
Conclusion:-
In parallelogram ABCD; Δ APB ≅ Δ CQD
Question 7.
In and DEF, AB || DE, AB=DE; BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that
(i) ABED is a parallelogram
(ii) BCFE is a parallelogram
(iii) AC = DF
(iv) ∆ABC ≅ ∆DEF
Answer:
Given:- AB||DE , AB=DE; BC||EF, BC=EF
Formula used:- SSS congruency rule
If all 3 sides of both triangle are equal then
Both triangles are congruent.
Solution:-
(1) In ABED
AB||DE
AB=DE
∴ ABED is a parallelogram
∵ If one pair of side in quadrilateral is equal and parallel
Then the quadrilateral is parallelogram.
(2) In BCFE
BC||EF
BC=EF
∴ BCEF is a parallelogram
∵ If one pair of side in quadrilateral is equal and parallel
Then the quadrilateral is parallelogram.
(3) As ABED is a parallelogram
BE||AD , BE=AD ;
As BCFE is a parallelogram
BE||CF , BE=CF ;
∴ By concluding both above statements
AD||CF and AD=CF
∴ ACFD is a parallelogram
∵ If one pair of side in quadrilateral is equal and parallel
Then the quadrilateral is parallelogram
If ACDF is a parallelogram ;
Then
AC=DF [opposite sides of parallelogram are equal ]
(4) In Δ ABC and Δ DEF
AB=DE [Given]
BC=EF [Given]
AC=DF [Proved above]
Δ ABC ≅ Δ DEF [SSS congruency rule]
Question 8.
ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are the points of tri section of the diagonal BD. Prove that CQ || AP and also AC bisects PQ (see figure).
Answer:
Given:- ABCD is parallelogram
BP=BD/3
DQ=BD/3
Formula used:- SAS congruency property
If 2 sides and angle between the two sides of both
the triangle are equal, then both triangle are congruent
Solution:-
⇒ In parallelogram ABCD
AO=OC and DO=OB [diagonal of parallelogram bisect each other]
As DO=OB
Where DO=DQ+OQ
OB=OP+PB
∴ DQ+OQ=OP+PB
⇒ +OQ=OP+
⇒ OQ=OP+ -
⇒ OQ=OP
PQ = OP+OQ = 2(OQ) = 2(OP)
∴ AC diagonal bisect PQ
⇒ In Δ QOC and Δ POA
OQ=PO [Proven above]
AO=OC [Diagonal of parallelogram bisect each other]
∠ QOC=∠ AOP [vertically opposite angles]
Hence Δ QOC ≅ Δ POA
∴ ∠ CQO=∠ OPA
If QC and PA are 2 lines
And QP is the transverse
And ∠ CQO=∠ OPA by Alternate angles
∴ QC||PA
Conclusion:-
CQ||PA and CA is bisector of PQ.
Question 9.
ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.
Answer:
Given:- *ABCD is a square
*E, F, G and H are the mid points of AB, BC, CD and DA
respectively
*AE = BF = CG = DH
Formula used:- *Isosceles Δ property
⇒ If 2 sides of triangle are equal then
Corresponding angle will also be equal
*Properties of quadrilateral to be square
⇒ All sides are equal
⇒ All angles are 90°
Solution:-
In Δ AHE, Δ EBF, Δ FCG, Δ DHG
⇒ AE = BF = CG = DH [Given]
If;
⇒ AE=EB [E is midpoint of AB]
⇒ BF=FC [F is midpoint of BC]
⇒ CG=GD [G is midpoint of CD]
⇒ DH=HA [H is midpoint of DA]
⇒ AE = BF = CG = DH
On replacing every part we get;
⇒ EB=FC=GD=HA;
⇒ ∠ A+∠ B+∠ C+∠ D=90° [All angles of square are 90°]
Hence;
All triangles Δ AHE, Δ EBF, Δ FCG, Δ DHG
are congruent by SAS property
∴ Δ AHE≅ Δ EBF≅ Δ FCG ≅ Δ DHG
⇒ HE=EF=FG=GH [All triangles are congruent]
In Δ AHE, Δ EBF, Δ FCG, Δ DHG
∵ all sides of square are equal and after the midpoint of each sides
Every half side of square are equal to half of other sides.
HA=AE , EB=FB ,FC=GC ,HD=DG
∴ All Δ AHE, Δ EBF, Δ FCG, Δ DHG are isosceles
⇒ as central angle of all triangle is 90°
It makes all Δ AHE, Δ EBF, Δ FCG, Δ DHG are right angle isosceles Δ
∴ all corresponding angles of equal side will be 45°
∠ AHE=∠ BEF=∠ CFG=∠ DHG=∠ AEH=∠ BFE=∠ CGF=∠ DGH=45°
⇒ as AB is straight line
Then; ∠ AEH+∠ HEF+∠ BEF=180°
45°+∠ HEF+45° =180°
∠ HEF=180° -90° =90°
Similarly ;
∠ EFG=90°
∠ FGH=90°
∠ GHE=90°
Conclusion:-
All angles are 90° and all sides are equal of quadrilateral
Hence quadrilateral is square
Exercise 8.4
Question 1.ABC is a triangle. D is a point on AB such that 
and E is a point on AC such that 
If DE = 2 cm find BC.
Answer:
Given:- DE=2cm
Formula used:- Line joining midpoints of 2 sides of triangle will be
half of its 3rd side.
Solution:-
Assume mid points of line AB and AC is M and N
The formation of new triangle will be there Δ AMN
As;
⇒ AC=4AE
and 
∴ AM=2AE
Hence;
E is the midpoint of AM
Similarly
D is the midpoint of AN
→ In Δ AMN
If E,D are midpoints
ED=
MN
MN=2ED
MN=2×2cm=4cm
→ In Δ ABC
If M,N are midpoints
AB=MN
AB=2MN
AB=2×4cm=8cm
Conclusion:-
BC=8cm
Question 2.ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.
Answer:
Given:- ABCD is quadrilateral E, F, G and H are the midpoints of
AB, BC, CD and DA respectively
Formula used:- Line joining midpoints of 2 sides of triangle
Is parallel to 3rd side
Solution:-
BD is diagonal of quadrilateral
EH is the line joined by midpoints of triangle ABD,
∴EH is parallel to BD
GF is line joined by midpoints of side BC&BD of triangle BCD
∴GF is parallel to BD
⇒ If HE is parallel to BD and BD is parallel to GF
∴ It gives HE is parallel to GF
⇒ AC is another diagonal of quadrilateral
GH is the line joined by midpoints of triangle ADC,
∴GH is parallel to AC
FE is line joined by midpoints of side BC&AB of triangle ABC
∴FE is parallel to AC
⇒ If GH is parallel to AC and AC is parallel to FE
∴ It gives GH is parallel to FE
As HE||GF and GH||FE
∴ EFGH is a parallelogram
Conclusion:-
EFGH is a parallelogram
Question 3.Show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle.
Answer:
Given:- ABCD is a rhombus
E,F,G,H are the mid points of AB,BC,CD,DA
Formula used:- Line joining midpoints of 2 sides of triangle
Is parallel and half of 3rd side
Solution:-
BD is diagonal of rhombus
EH is the line joined by midpoints of triangle ABD,
∴EH is parallel and half of BD
GF is line joined by midpoints of side BC&BD of triangle BCD
∴GF is parallel and half BD
⇒ If HE is parallel to BD and BD is parallel to GF
∴ It gives HE is parallel to GF
⇒ If HE is half of BD and GF is also half of BD
∴ It gives HE is equal to GF
⇒ AC is another diagonal of rhombus
GH is the line joined by midpoints of triangle ADC,
∴GH is parallel to AC
∴GH is half of AC
FE is line joined by midpoints of side BC&AB of triangle ABC
∴FE is parallel to AC
∴ FE is half of AC
⇒ If GH is parallel to AC and AC is parallel to FE
∴ It gives GH is parallel to FE
⇒ If GH is half of AC and FE is also half of AC
∴ It gives GH is equal to FE
As diagonals of rhombus intersect at 90°
And AC is parallel to FE and GH
All angles of EFGH is 90°
All angles are 90° and opposite sides are equal and parallel
Conclusion:-
EFGH is a rectangle
Question 4.In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC respectively. Show that the line segments AF and EC trisect the diagonal BD.
Answer:Given:- ABCD is a parallelogram
E and F are the midpoints of the sides AB and DC respectively
Formula used:- Line drawn through midpoint of one side of triangle
Parallel to other side , bisect the 3rd side.
Solution:-
As ABCD is parallelogram
AB=CD and AB||CD;
⇒ AE=CF and AE||CF [E and F are the midpoints of AB and CD]
In quadrilateral AECF
⇒ AE=CF and AE||CF
AECF is a parallelogram.
∵ Quadrilateral having one pair of side equal and parallel are parallelogram
If AECF is a parallelogram
∴ AF||CE
Hence ;
PF||CQ and AP||QE [As AF=AP+PF and CE=CQ+QE ]
In Δ DQC
DF=FC [F is midpoint]
PF||CQ
Then;
P is midpoint of DQ
DP=PQ
∵ Line drawn through midpoint of one side of triangle Parallel to
other side , bisect the 3rd side
In Δ APB
AE=EB [E is midpoint]
AP||QE
Then;
Q is midpoint of PB
PQ=QB
∵ Line drawn through midpoint of one side of triangle Parallel to
other side , bisect the 3rd side
If DP=PQ and PQ=QB
Then DP=PQ=QB
Conclusion:-
Line segments AF and EC trisect the diagonal BD.
Question 5.Show that the line segments joining the midpoints of the opposite sides of a quadrilateral and bisect each other.
Answer:
Given:- ABCD is a quadrilateral
Formula used:- Line joining midpoints of 2 sides of triangle
Is parallel and half of 3rd side
Solution:-
BD is diagonal of quadrilateral
EH is the line joined by midpoints of triangle ABD,
∴EH is parallel and half of BD
GF is line joined by midpoints of side BC&BD of triangle BCD
∴GF is parallel and half BD
⇒ If HE is parallel to BD and BD is parallel to GF
∴ It gives HE is parallel to GF
⇒ If HE is half of BD and GF is also half of BD
∴ It gives HE is equal to GF
⇒ AC is another diagonal of quadrilateral
GH is the line joined by midpoints of triangle ADC,
∴GH is parallel to AC
∴GH is half of AC
FE is line joined by midpoints of side BC&AB of triangle ABC
∴FE is parallel to AC
∴ FE is half of AC
⇒ If GH is parallel to AC and AC is parallel to FE
∴ It gives GH is parallel to FE
⇒ If GH is half of AC and FE is also half of AC
∴ It gives GH is equal to FE
If both opposite sides are parallel and equal
Then, the quadrilateral is parallelogram
If EFGH is parallelogram
Then their diagonal bisect each other
If diagonal of parallelogram is the line joining midpoint of opposite sides of quadrilateral
Then;
Line joining midpoints of opposite sides of quadrilateral bisect each other.
Conclusion:-
Lines joining midpoints of opposite sides of quadrilateral bisect each other
Question 6.ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and Parallel to BC intersects AC at D. Show that
(i) D is the midpoint of AC
(ii) MD ⊥ AC
(iii) CM = MA = 
AB
Answer:Given:- ACB is right angle triangle
M is midpoint of AB
DM||CB
Formula used:- Line drawn through midpoint of one side of triangle
Parallel to other side , bisect the 3rd side
* SAS congruency property
If 2 sides and angle between the two sides of both
the triangle are equal, then both triangle are congruent
Solution:-
(1) In Δ ABC
M is midpoint of AB
And DM||CB
∴ D is midpoint of AC
AD=DC
∵ Line drawn through midpoint of one side of triangle
Parallel to other side , bisect the 3rd side
(2) In Δ ABC
DM||CB
∠ ADM=∠ ACB [Corresponding angles]
∠ ACB =90°
∠ ADM=90°
(3) In Δ ADM and Δ DMC
⇒ DM=DM [Common in both triangles ]
⇒ AD=DC [D is the midpoint]
As ADC is straight line
∠ ADM+∠ MDC=180°
∠ MDC=180° - ∠ ADM
∠ MDC=90°
⇒ ∠ ADM=∠ MDC
Hence;
In Δ ADM ≅ Δ DMC
∴ CM=MA
⇒ MA=AB [M is midpoint of AB]
∴ CM=MA=AB
ABC is a triangle. D is a point on AB such that and E is a point on AC such that If DE = 2 cm find BC.
Answer:
Given:- DE=2cm
Formula used:- Line joining midpoints of 2 sides of triangle will be
half of its 3rd side.
Solution:-
Assume mid points of line AB and AC is M and N
The formation of new triangle will be there Δ AMN
As;
⇒ AC=4AE
and
∴ AM=2AE
Hence;
E is the midpoint of AM
Similarly
D is the midpoint of AN
→ In Δ AMN
If E,D are midpoints
ED=MN
MN=2ED
MN=2×2cm=4cm
→ In Δ ABC
If M,N are midpoints
AB=MN
AB=2MN
AB=2×4cm=8cm
Conclusion:-
BC=8cm
Question 2.
ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.
Answer:
Given:- ABCD is quadrilateral E, F, G and H are the midpoints of
AB, BC, CD and DA respectively
Formula used:- Line joining midpoints of 2 sides of triangle
Is parallel to 3rd side
Solution:-
BD is diagonal of quadrilateral
EH is the line joined by midpoints of triangle ABD,
∴EH is parallel to BD
GF is line joined by midpoints of side BC&BD of triangle BCD
∴GF is parallel to BD
⇒ If HE is parallel to BD and BD is parallel to GF
∴ It gives HE is parallel to GF
⇒ AC is another diagonal of quadrilateral
GH is the line joined by midpoints of triangle ADC,
∴GH is parallel to AC
FE is line joined by midpoints of side BC&AB of triangle ABC
∴FE is parallel to AC
⇒ If GH is parallel to AC and AC is parallel to FE
∴ It gives GH is parallel to FE
As HE||GF and GH||FE
∴ EFGH is a parallelogram
Conclusion:-
EFGH is a parallelogram
Question 3.
Show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle.
Answer:
Given:- ABCD is a rhombus
E,F,G,H are the mid points of AB,BC,CD,DA
Formula used:- Line joining midpoints of 2 sides of triangle
Is parallel and half of 3rd side
Solution:-
BD is diagonal of rhombus
EH is the line joined by midpoints of triangle ABD,
∴EH is parallel and half of BD
GF is line joined by midpoints of side BC&BD of triangle BCD
∴GF is parallel and half BD
⇒ If HE is parallel to BD and BD is parallel to GF
∴ It gives HE is parallel to GF
⇒ If HE is half of BD and GF is also half of BD
∴ It gives HE is equal to GF
⇒ AC is another diagonal of rhombus
GH is the line joined by midpoints of triangle ADC,
∴GH is parallel to AC
∴GH is half of AC
FE is line joined by midpoints of side BC&AB of triangle ABC
∴FE is parallel to AC
∴ FE is half of AC
⇒ If GH is parallel to AC and AC is parallel to FE
∴ It gives GH is parallel to FE
⇒ If GH is half of AC and FE is also half of AC
∴ It gives GH is equal to FE
As diagonals of rhombus intersect at 90°
And AC is parallel to FE and GH
All angles of EFGH is 90°
All angles are 90° and opposite sides are equal and parallel
Conclusion:-
EFGH is a rectangle
Question 4.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC respectively. Show that the line segments AF and EC trisect the diagonal BD.
Answer:
Given:- ABCD is a parallelogram
E and F are the midpoints of the sides AB and DC respectively
Formula used:- Line drawn through midpoint of one side of triangle
Parallel to other side , bisect the 3rd side.
Solution:-
As ABCD is parallelogram
AB=CD and AB||CD;
⇒ AE=CF and AE||CF [E and F are the midpoints of AB and CD]
In quadrilateral AECF
⇒ AE=CF and AE||CF
AECF is a parallelogram.
∵ Quadrilateral having one pair of side equal and parallel are parallelogram
If AECF is a parallelogram
∴ AF||CE
Hence ;
PF||CQ and AP||QE [As AF=AP+PF and CE=CQ+QE ]
In Δ DQC
DF=FC [F is midpoint]
PF||CQ
Then;
P is midpoint of DQ
DP=PQ
∵ Line drawn through midpoint of one side of triangle Parallel to
other side , bisect the 3rd side
In Δ APB
AE=EB [E is midpoint]
AP||QE
Then;
Q is midpoint of PB
PQ=QB
∵ Line drawn through midpoint of one side of triangle Parallel to
other side , bisect the 3rd side
If DP=PQ and PQ=QB
Then DP=PQ=QB
Conclusion:-
Line segments AF and EC trisect the diagonal BD.
Question 5.
Show that the line segments joining the midpoints of the opposite sides of a quadrilateral and bisect each other.
Answer:
Given:- ABCD is a quadrilateral
Formula used:- Line joining midpoints of 2 sides of triangle
Is parallel and half of 3rd side
Solution:-
BD is diagonal of quadrilateral
EH is the line joined by midpoints of triangle ABD,
∴EH is parallel and half of BD
GF is line joined by midpoints of side BC&BD of triangle BCD
∴GF is parallel and half BD
⇒ If HE is parallel to BD and BD is parallel to GF
∴ It gives HE is parallel to GF
⇒ If HE is half of BD and GF is also half of BD
∴ It gives HE is equal to GF
⇒ AC is another diagonal of quadrilateral
GH is the line joined by midpoints of triangle ADC,
∴GH is parallel to AC
∴GH is half of AC
FE is line joined by midpoints of side BC&AB of triangle ABC
∴FE is parallel to AC
∴ FE is half of AC
⇒ If GH is parallel to AC and AC is parallel to FE
∴ It gives GH is parallel to FE
⇒ If GH is half of AC and FE is also half of AC
∴ It gives GH is equal to FE
If both opposite sides are parallel and equal
Then, the quadrilateral is parallelogram
If EFGH is parallelogram
Then their diagonal bisect each other
If diagonal of parallelogram is the line joining midpoint of opposite sides of quadrilateral
Then;
Line joining midpoints of opposite sides of quadrilateral bisect each other.
Conclusion:-
Lines joining midpoints of opposite sides of quadrilateral bisect each other
Question 6.
ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and Parallel to BC intersects AC at D. Show that
(i) D is the midpoint of AC
(ii) MD ⊥ AC
(iii) CM = MA = AB
Answer:
Given:- ACB is right angle triangle
M is midpoint of AB
DM||CB
Formula used:- Line drawn through midpoint of one side of triangle
Parallel to other side , bisect the 3rd side
* SAS congruency property
If 2 sides and angle between the two sides of both
the triangle are equal, then both triangle are congruent
Solution:-
(1) In Δ ABC
M is midpoint of AB
And DM||CB
∴ D is midpoint of AC
AD=DC
∵ Line drawn through midpoint of one side of triangle
Parallel to other side , bisect the 3rd side
(2) In Δ ABC
DM||CB
∠ ADM=∠ ACB [Corresponding angles]
∠ ACB =90°
∠ ADM=90°
(3) In Δ ADM and Δ DMC
⇒ DM=DM [Common in both triangles ]
⇒ AD=DC [D is the midpoint]
As ADC is straight line
∠ ADM+∠ MDC=180°
∠ MDC=180° - ∠ ADM
∠ MDC=90°
⇒ ∠ ADM=∠ MDC
Hence;
In Δ ADM ≅ Δ DMC
∴ CM=MA
⇒ MA=AB [M is midpoint of AB]
∴ CM=MA=AB
