Question

In the following figure, if AC = BE, then prove AD = CE.
(3 marks)

Answer 1

From figure, AC = AB.
So, $\angle ACB=\angle ABC\cdots \left(I\right)$ [angles opposite to equal sides are equal]
and $\angle CBE+\angle ABC={180}^{\circ }$ [linear pair].
$\Rightarrow \angle CBE={180}^{\circ }-\angle ABC\cdots \left(II\right)$
Similarly, $\angle ACD+\angle ACB={180}^{\circ }$ (Linear pair)
$\therefore \angle ACD={180}^{\circ }-\angle ACB-\left(III\right)$
But from $\left(I\right)$, $\angle ACB=\angle ABC$
$\Rightarrow$ $\angle ACD=\angle CBE$ (from $\left(II\right)$ and $\left(III\right)$)

(1 mark)

In $\Delta ACD$ and $\Delta EBC,$
AC = EB (given)
$\angle ACD=\angle EBC$ (Proved above)
CD = BC (Given)

(1 mark)

So, by SAS postulate,
$\Delta ACD\cong \Delta EBC$
Hence, AD = EC (CPCT)

(1 mark)