Given: ∆ ABC ∼ ∆ PQR
To Prove:
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A(∆ ABC)
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=
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AB2
|
=
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BC2
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=
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AC2
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||
A(∆ PQR)
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PQ2
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QR2
|
PR2
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Construction: Draw seg AM ┴ side BC and seg PN ┴ side QR.
Solution . ∆ ABC ∼
∆ PQR
∴
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AB
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=
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BC
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=
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AC
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[C.S.S.T.]--------------------eq.
(1)
|
|
PQ
|
QR
|
PR
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and ∠ ABC ≅ ∠ PQR [c.a.s.t.]
i.e. ∠
ABM ≅ ∠ PQN ---------------------eq. (2)
[Same angle with different Names]
In ∆ ABM and ∆ PQN,
∠ ABM ≅ ∠ PQN [From
(2) ]
∠ AMB ≅ ∠ PNQ (Both are 900)
∴ ∆ AMB ∼ ∆ PNQ
∴
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AM
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=
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MB
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=
|
AB
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--------------------eq.
(3)
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PN
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NQ
|
PQ
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Now,
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A(∆ ABC)
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=
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½ (Base)(Height)
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||||
A(∆ PQR)
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½ (Base)(Height)
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||||||
∴
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A(∆ ABC)
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=
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BC × AM
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||||
A(∆ PQR)
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QR × PN
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||||||
∴
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A(∆ ABC)
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=
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AB × AB
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[From eq. (1) and
(3)]
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|||
A(∆ PQR)
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PQ × PQ
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