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(Image 1)

Given: △ABC & △PQR

AD is the median of △ABC

PM is the median of △PQR

PQAB=PRAC=PMAD→1

To prove: △ABC∼△PQR

Proof:Let us extend AD to point D such that that AD=DE and PM upto point L such that PM=ML

Join B to E, C to E, & Q to L and R to L

(Image 2)

We know that medians is the bisector of opposite side

Hence BD=DC & AD=DE *By construction)

Hence in quadrilateral ABEC, diagonals AE and BC bisect each other at point D

∴ABEC is a parallelogram

∴AC=BE & AB=EC (opposite sides of a parallelogram are equal) →2

Similarly we can prove that

PQLR is a parallelogram.

PR=QL,PQ=LR (opposite sides of a parallelogram are equal) →3

Given that

PQAB=PRAC=PMAD (frim 1)

⇒PQAB=QLBE=PMAD  (from 2 and 3)

⇒PQAB=QLBE=2PM2A