Question

$△ABC$ is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. What is the measure of $\angle BCD$?

Answer 1

Given, AB = AC and BA = AD

In $△ABC$,

$\angle ABC=\angle ACB=x$ (opposite angles of equal sides)

In $△ACD$,

$\angle ADC=\angle ACD=y$ (opposite angles of equal sides) (1 mark)

In $△BCD$,

$\angle DBC+\angle BCD+\angle CDB={180}^{\circ }$

$x+x+y+y={180}^{\circ }$

$2x+2y={180}^{\circ }$

$x+y={90}^{\circ }$

$\angle BCD=(x+y)={90}^{\circ }$

$\angle BCD={90}^{\circ }$ (1 mark)