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Let ABC be a triangle with \angle C = 90^\circ and |CA| \ne |CB|. Let \overline{CH} be an altitude and \overline{CL} be an interior angle bisector. Show that for X \ne C on the line \overline{CL}, we have \angle XAC \ne \angle XBC. Also show that for Y \ne C on the line \overline{CH} we have \angle YAC \ne \angle YBC.

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