Question

In the following figure if AC =EB, then the side AD is congruent to

• A. BD
• B. CE
• C. AE
• D. BE

Answer 1

The correct option is B

CE

$\Delta ABC$ is isosceles.

So, $\angle ACB=\angle ABC-\left(I\right)$

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

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

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

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

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

$\angle ACD=\angle CBE.$

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.]