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Question Number 20545 by Tinkutara last updated on 23/Sep/17

Let ABC be an acute-angled triangle with AC ≠ BC and let O be the circumcenter and F be the foot of altitude through C. Further, let X and Y be the feet of perpendiculars dropped from A and B respectively to (the extension of) CO. Prove that FY ⊥ CA using that ∠CFY = ∠CBY = ∠CAF.

$$\mathrm{Let}\:\mathrm{ABC}\:\mathrm{be}\:\mathrm{an}\:\mathrm{acute}-\mathrm{angled}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\mathrm{AC}\:\neq\:\mathrm{BC}\:\mathrm{and}\:\mathrm{let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{circumcenter}\:\mathrm{and}\:\mathrm{F}\:\mathrm{be}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of} \\ $$$$\mathrm{altitude}\:\mathrm{through}\:\mathrm{C}.\:\mathrm{Further},\:\mathrm{let}\:\mathrm{X}\:\mathrm{and} \\ $$$$\mathrm{Y}\:\mathrm{be}\:\mathrm{the}\:\mathrm{feet}\:\mathrm{of}\:\mathrm{perpendiculars}\:\mathrm{dropped} \\ $$$$\mathrm{from}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{respectively}\:\mathrm{to}\:\left(\mathrm{the}\right. \\ $$$$\left.\mathrm{extension}\:\mathrm{of}\right)\:\mathrm{CO}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{FY}\:\bot\:\mathrm{CA} \\ $$$$\mathrm{using}\:\mathrm{that}\:\angle\mathrm{CFY}\:=\:\angle\mathrm{CBY}\:=\:\angle\mathrm{CAF}. \\ $$

Commented by Tinkutara last updated on 23/Sep/17

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