Question

In the following figure if AC = AB = EB, and DC = BC, then prove that the side AD is congruent to EC. $\left[3\right]$

Answer 1

$\Delta ABC$ is isosceles.

So, $\angle ACB=\angle ABC$ ____(I)

$\angle CBE+\angle ABC={180}^{\circ }$ [linear pair]

$\Rightarrow \angle CBE={180}^{\circ }-\angle ABC$ ____(II)

Similarly, $\angle DCA+\angle ACB={180}^{\circ }$ (Linear pair)

$\therefore \angle ACD={180}^{\circ }-\angle ACB$ ____(III)

From (I), (II), and (III), we get

$\angle ACD=\angle CBE.$ $\left[1.5marks\right]$

In $\Delta ACD$ and $\Delta EBC,$

$AC=EB$ (Given)

$\angle ACD=\angle EBC$ (Proved above)

$CD=BC$ (Given)

So, by SAS postulate, $\Delta ACD\cong \Delta EBC$

Hence, $AD=EC$ [C.P.C.T.C.] $\left[1.5marks\right]$