Question

Sides AB and AC and median AD of a
triangle ABC are respectively
proportional to sides PQ and PR and
median PM of another triangle PQR.
Show that A ABC ~ APQR.​

Answer 1

Step-by-step explanation:

Given: △ABC & △PQR

AD is the median of △ABC

PM is the median of △PQR

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

We know that medians is the bisector of opposite side

Hence BD=DC & AD=DE *By construction)→1

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

PQ/AB = PR/AC= PM/AD(frim 1)

⇒PQ/AB = QL/BE= PM/AB(from 2 and 3)

⇒ PQ/AB= QL/BE= 2PM/2AD

⇒ PQ/AB= QL/BE= PL/AE

(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→4

Similarly we can prove that

△AEC∼△PLR

We know that corresponding angles of similar triangles are equal

∠CAE=∠RPL→5

Adding 4 and 5, we get

∠BAE+∠CAE=∠QPL+∠RPL

⇒∠CAB=∠RPQ→6

In △ABC and △PQR

PQ/AB= PR/AC(from 1)

∠CAB=∠RPQ (from 6)

∴△ABC∼△PQR (By SAS similarity criteria)

Answer 2

hope it's help u..........