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Show that for X <> C on line CL, <XAC = <XBC

Source: JBMO 2001, Problem 2

October 30, 2005
geometrycircumcirclesymmetrytrigonometryperpendicular bisectorangle bisectorpower of a point

Problem Statement

Let ABCABC be a triangle with ∠C=90∘\angle C = 90^\circ and CA≠CBCA \neq CB. Let CHCH be an altitude and CLCL be an interior angle bisector. Show that for X≠CX \neq C on the line CLCL, we have ∠XAC≠∠XBC\angle XAC \neq \angle XBC. Also show that for Y≠CY \neq C on the line CHCH we have ∠YAC≠∠YBC\angle YAC \neq \angle YBC. Bulgaria
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