Question 43
In $$\triangle ABC$$, the perpendiculars drawn from $$A, B$$ and $$C$$ meet the opposite sides at $$D, E$$ and $$F$$, respectively. $$AD, BE$$ and $$CF$$ intersect at point $$P$$. If $$\angle EPD = 116^\circ$$ and the bisectors of $$\angle A$$ and $$\angle B$$ meet at $$Q$$, then the measure of $$\angle AQB$$ is:
Solution
In quadrilateral EPDC,
$$\angle PDC +\angle DCE +Â \angle CEP +Â \angle EPD = 360$$
$$\angle DCE = 360 - 90 - 90 - 116$$
$$\angle DCE = 64\degree$$
$$\angle AQB = 90 + \frac{\angle DCE }{2}$$
= $$90 + \frac{64}{2}$$ = 90 + 32 = $$122\degree$$
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