NECO 2023 · Paper 1 · Q39

Angles, triangles & polygons

$ABC$ is an isosceles triangle, and $E$ and $D$ are points on $AC$ and $BC$ respectively such that $BE \perp AC$ and $ED \perp BC$ . If $\angle ABE = 68^\circ$ and $\angle A = \angle C$ , find $\angle CED$ .

Worked solution (try it first)

$BE \perp AC$ , so triangle $ABE$ has a right angle at $E$ : $\angle A = 90^\circ - 68^\circ = 22^\circ$ .
$\angle C = \angle A = 22^\circ$ .
$ED \perp BC$ , so triangle $CDE$ has a right angle at $D$ : $\angle CED = 90^\circ - 22^\circ$
$= 68^\circ$ , option E.

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