In the acute-angled triangle ABC, the foot of the perpendicular from
B to CA is E. Let l be the tangent to the circle ABC at B. The foot
of the perpendicular from C to l is F. Prove that EF is parallel to
AB.

Can someone correct me? I have a proof based on the the circumcentre of the triangle. Yet I am unsure whether the question describes the orthocentre of circumcentre?

The question as stated is fine. Just use the fact that the inscribed angle is equal to the angle between the tangent and the chord that is created after joining the endpoints of the inscribed angle. \(BECF\) is cyclic, thus there's another equal angle. Now use the property of tangent lines (some angles equal \(\implies\) the lines are tangent).

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TopNewestCan someone correct me? I have a proof based on the the circumcentre of the triangle. Yet I am unsure whether the question describes the orthocentre of circumcentre?

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The question as stated is fine. Just use the fact that the inscribed angle is equal to the angle between the tangent and the chord that is created after joining the endpoints of the inscribed angle. \(BECF\) is cyclic, thus there's another equal angle. Now use the property of tangent lines (some angles equal \(\implies\) the lines are tangent).

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