JAMB 1990 · UME · Q34

In triangles $XYZ$ and $XQP$ , $XP = 4$ cm, $XQ = 5$ cm and $PQ = QY = 3$ cm, with $\angle XQP = \angle XZY$ . Find $ZY$ .

Worked solution (try it first)

Triangles $XQP$ and $XZY$ share $\angle X$ and have $\angle XQP = \angle XZY$ , so they are similar, with $Q$ matching $Z$ and $P$ matching $Y$ .
So $\dfrac{PQ}{YZ} = \dfrac{XP}{XY}$ .
$Q$ lies on $XY$ , so $XY = XQ + QY = 5 + 3 = 8$ cm.
$\dfrac{3}{ZY} = \dfrac{4}{8}$ , so $ZY = 6$ cm, option B.