Question

Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians

Given :ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q.

AP= 1 CM, BP= 3 CM, AQ=1.5 CM, QC=4.5

IF YOU ANSWER I WILL MARK YOU AS BRAINLIST PLS

Answer 1

consider two triangles ABC and PQR in which ad and pm are medians respectively.

given : triangle ABC ~ triangle PQR

ANGLE A = ANGLE P

ANGLE B = ANGLE Q

ANGLE C = ANGLE R

SO , AB/PQ, BC/ QR , AC /PR

RTP: ar( ABC)/ ar ( PQR) = (AD/PM)^2

PROOF : IN ABD AND PQM

AB/PQ= BC/QR ( given)

AB / PQ = I/2 BC/ 1/2 QR

AB/PQ = BD/ QM

ANGLE B = ANGLE Q

BY SAS SIMILARITY

triangle ABD~ TRIANGLE PQM

SO, AB/PQ =BD/QM=AD/PM

WE KNOW THAT

ar(ABC)/ar(PQR)= (AB/PQ)^2=(BC/QR)^2=(AC/PR)^2

ar(ABC)/ar(PQR) = (AB/PQ)^2

AB/PQ = AD/PM

SO, ar(ABC)/ ar(PQR)= (AD/PM)^2