Let ABC be an acute angled triangle. The circle Γ with BC as diameter intersects AB and AC again at P and Q, respectively.
Let ABC be an acute angled triangle. The circle Γ with BC as diameter intersects AB and AC again at P and Q, respectively. Determine ∠BAC given that the orthocenter of triangle APQ lies on Γ.
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Let K denote the orthocenter of triangle APQ. Since triangles ABC and AQP are similar it follows that K lies in the interior of triangle APQ.
Note that ∠KPA = ∠KQA = 900 - ∠A. Since BPKQ is a cyclic quadrilateral it follows that ∠BQK = 1800 - ∠BPK = 900 - ∠A, while on the other hand ∠BQK = ∠BQA - ∠KQA = ∠A since BQ is perpendicular to AC. This shows that 900 - ∠A = ∠A, so ∠A = 450.
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