Question

In the figure below, $\angle$C = ${40}^{\circ }$ and CB = AC. Also, EC is the bisector of AB. Calculate $\angle$AC in degrees.

Answer 1

Given, $\angle C={40}^{\circ }$, CB = AC and EC is the bisector of AB.
We need to find the value of $\angle$ACE.

In $△$AEC and $△$BEC,

BC = AC (Given)

$\angle$A=$\angle$B
[$\because$ CB = AC and angles opposite to equal sides are equal]

AE=BE (EC is the bisector of AB)

$△$AEC $\cong$$△$BEC (SAS congruency)

$\angle$BCE = $\angle$ACE (CPCTE)

$\angle$C =$\angle$BCE + $\angle$ACE = 2$\angle$ACE

$\angle$ACE = $\frac{\angle C}{2}$= $\frac{{40}^{\circ }}{2}$ = ${20}^{\circ }$