Question

In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE, Prove that segment DE || BC.

Answer 1

Side BA and CA of $\Delta ABC$ are produced such that BA = AD are CA = AE. ED is joined.

To prove : DE || BC

Proof : In $\Delta ABCand\Delta DAE$

AB = AD (Given)

AC = AE ( Given)

$\angle BAC=\angle DAE$ (Vertically opposite angles )

$\therefore \Delta ABC\cong \Delta DAE$ (SAS axiom)

$\therefore \angle ABC=\angle ADE$ (c.p.c.t)

But these are alternate angles.

$\therefore$ DE || BC