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WAEC Mathematics 2025 Paper

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Question : 39

Total: 50

In the diagram below, AB||DC, the bisectors of < BAC and < ACD meet at E. Find the value of < AEC

45°
60°
30°
90°

Solution:

Concept:

Since

AB \parallel DC

and

AC

is a transversal, the co-interior angles add up to

180^\circ

. Then use the angle sum of a triangle.

Explanation:

Given

AB \parallel DC

and

AC

intersects them at

A

and

C

.

Therefore,

\angle BAC + \angle ACD = 180^\circ

(co-interior angles).

AE bisects

\angle BAC

, so

\angle EAC = \frac{1}{2}\angle BAC

.

CE bisects

\angle ACD

, so

\angle ACE = \frac{1}{2}\angle ACD

.

Thus,

\angle EAC + \angle ACE = \frac{1}{2}(\angle BAC + \angle ACD) = \frac{1}{2}(180^\circ) = 90^\circ

.

In

\triangle AEC

,

\angle EAC + \angle ACE + \angle AEC = 180^\circ

.

Hence,

90^\circ + \angle AEC = 180^\circ

.

So,

\angle AEC = 90^\circ

.

Answer:

\angle AEC = 90^\circ

, i.e. Option D.

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