: △ABC & △PQRAD is the median of △ABCPM is the median of △PQRPQAB = PRAC = PMAD →1To prove: △ABC∼△PQRProof:Let us extend AD to point D such that that AD=DE and PM upto point L such that PM=MLJoin 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) →2Similarly we can prove that PQLR is a parallelogram.PR=QL,PQ=LR (opposite sides of a parallelogram are equal) →3Given that PQAB = PRAC = PMAD (frim 1)⇒ PQAB = QLBE = PMAD (from 2 and 3)⇒ PQAB = QLBE = 2PM2AD ⇒ PQAB = QLBE = PLAE (As AD=DE,AE=AD+DE=AD+AD=2AD & PM=ML,PL=PM+ML=PM+PM=2PM)∴△ABE∼△PQL (By SSS SIMILARITY criteria)We know that corresponding angles of similar triangles are equal∴∠BAE=∠QPL→4Similarly we can prove that△AEC∼△PLRWe know that corresponding angles of similar triangles are equal∠CAE=∠RPL→5Adding 4 and 5, we get∠BAE+∠CAE=∠QPL+∠RPL⇒∠CAB=∠RPQ→6In △ABC and △PQRPQAB = PRAC (from 1)∠CAB=∠RPQ (from 6)∴△ABC∼△PQR (By SAS similarity criteria)Hence, proved Step-by-step explanation:Hope it's is helpful for you