WAEC 2025 · Paper 1 · Q42✱✱

In $\triangle ABC$ , side $BC$ is produced to $D$ . If $|AB| = |AC|$ and $\angle BAC = 50^\circ$ , find $\angle ACD$ .

Worked solution (try it first)

$|AB| = |AC|$ , so the base angles at $B$ and $C$ are equal: each is $\frac{180^\circ - 50^\circ}{2} = 65^\circ$ .
$BCD$ is a straight line, so $\angle ACD = 180^\circ - 65^\circ$
$= 115^\circ$ , option A.