Question

In ABC PQ parallel BC and BD =DC Prove
that PE = EQ​

Answer 1

in ΔABD and ΔAPE

∠P =∠B

∠AEP=∠ADB

so by AA similarity

ΔABD and ΔAPE are similar

so AP/AB=AE/AD=PE/BD----eq(1)

again in ΔAEQ and ΔACD

∠AEQ=∠ADC

∠Q=∠C

so by AA similarity

ΔAEQ and ΔACD are similar

so AQ/AC=AE/AD=EQ/DC----eq(2)

from eq(1) and eq(2)

AE/AD=PE/BD=EQ/DC

PE/BD=EQ/DC

PE/EQ=BD/DC

given,BD=DC

so that PE/EQ=1

》》PE=EQ

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Answer 2

Answer:

In a triangle ABC PQ is parallel to BC

D is the midpoint of the triangle

to prove PE=QE

Proof in a triangle APE and ABD

angle AEP=angle ADB

angle APE=angle ABD

triangle APE ~triangle ABD

PE /BD= AE/AD. 1

similarly triangle AEQ~ ADC

(same as the above)

AE/AD=EQ/CD. 2

From 1 and 2 we get

PE/BD=EQ/CD

therefore PE=EQ