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Question

In an acute-angled triangle ABC with AB<AC, the circle Γ touches AB at B and passes through C intersecting AC again at D. Prove that the orthocentre of triangle ABD lies on the circle Γ if and only if it lies on the perpendicular bisector of BC.

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Solution

∠ADB=∠B∴△ADB and △ABC are similar. In particular, ABD is an acute-angled triangle. Let H denote the orthocenter of triangle ABD. ∴∠BHD=180∘−∠A .Suppose that H lies on Γ . Since AB < AC the point D lies on the segment AC and ∠C=180∘∠BHD=∠A . ∴BH is the perpendicular bisector of AC. Hence ∠HBC=∠ABC=∠HCB , so H lies on the perpendicular bisector of BC. Conversely, suppose that H lies on the perpendicular bisector of BC. Then ∠HCB=∠HBC=90∘∠C. Since ∠ABD=∠C it follows that ∠HDB=90∘∠C . Since ∠HCB=∠HDB we hae that H lies on Γ .

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