WAEC 2025 · Paper 1 · Q39✱✱

In the diagram, $AB \parallel CD$ , and the bisectors of $\angle BAC$ and $\angle ACD$ meet at $E$ . Find $\angle AEC$ .

Worked solution (try it first)

$AB \parallel CD$ , so $\angle BAC$ and $\angle ACD$ are co-interior and add up to $180^\circ$ .
$AE$ and $CE$ bisect them, so $\angle EAC + \angle ECA = 180^\circ \div 2$
$= 90^\circ$ .
The angles of triangle $AEC$ add up to $180^\circ$ : $\angle AEC = 180^\circ - 90^\circ$
$= 90^\circ$ , option D.