JAMB 1999 · UME · Q32

Circle geometry

In the diagram, $EFGH$ is a cyclic quadrilateral in which $EH \parallel FG$ , and $FH$ and $EG$ are chords. If $\angle FHG = 42^\circ$ and $\angle EFH = 34^\circ$ , calculate $\angle HEG$ .

Worked solution (try it first)

Call $\angle HFG = a$ .
$EH \parallel FG$ , so alternate angles give $\angle EHF = a$ , and angles in the same segment give $\angle EGF = \angle EHF = a$ .
Angles in the same segment: $\angle EGH = \angle EFH = 34^\circ$ .
So $\angle FGH = a + 34^\circ$ .
Triangle $FGH$ : $a + (a + 34^\circ) + 42^\circ = 180^\circ$ , so $2a = 104^\circ$ and $a = 52^\circ$ .
$\angle HEG$ and $\angle HFG$ both stand on arc $HG$ , so $\angle HEG = 52^\circ$ , option C.

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