WAEC 2025 · Paper 1 · Q41✱✱

In the diagram, $B$ lies on $AD$ , $C$ lies on $AE$ , and $DC$ meets $BE$ at $F$ . The angles marked $a^\circ$ are equal. Find $b + c$ .

Worked solution (try it first)

$BFE$ is a straight line, so $\angle BFC = 180^\circ - a$ (it sits next to the $a$ at $F$ ).
The angles of quadrilateral $ABFC$ add up to $360^\circ$ : $a + b + (180 - a) + c = 360$ .
The $a$ s cancel: $b + c + 180 = 360$ , so $b + c = 180^\circ$ , option C.