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




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




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


