Let ABC be an acute angled triangle. The circle Γ with BC as diameter

1.	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.
Answer» 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.

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.

Answer»

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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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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