If a line parallel to one side of a triangle intersects the other two sides in two distinct points, then the other two sides are divided in the same ratio by it.

Given: In ∆ PQR, Line MN || side QR, 

T o Prove:	PM	=	PN
	MQ		NR

Construction: Draw seg QN and seg RM.

Solution: ∆ PMN and ∆ QMN have equal heights with common vertex N

∴ A(∆ PMN) = PM ------------------ eq. no. (1) A(∆QMN) MQ

Similarly, ∆ PMN and ∆ RMN have equal height with common vertex M.

∴ A(∆ PMN) = PN ------------------------ eq. no. (2) A(∆ RMN) NR

Now, A (∆ QMN) = A(∆ RMN)  ----------------------- eq. no. (3) [ ∵ Both  the triangles are between two Parallel Lines and their bases is common]

From eq (1), (2) and (3) PM = PN MQ NR