Geometry
From a semicircle with diameter , 2 lines are drawn from to , intersecting the circle at and respectively. These lines meet at point . The orthocenter of the triangle lie on the circumference of the semicircle. Find .
The answer is 45.
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1 solution
. Let K denote the orthocenter of triangle AP Q. Since triangles ABC and AQP are similar it follows that K lies in the interior of triangle AP Q. Note that ∠KP A = ∠KQA = 90◦−∠A. Since BPKQ is a cyclic quadrilateral it follows that ∠BQK = 180◦ − ∠BPK = 90◦ − ∠A, while on the other hand ∠BQK = ∠BQA − ∠KQA = ∠A since BQ is perpendicular to AC. This shows that 90◦ − ∠A = ∠A, so ∠A = 45◦ .