Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it Ukraine

Let $\triangle ABC$ be an acute angled triangle and $\angle ACB = 60^\circ$ . Let $BL$ be a bisector line and $BH$ be an altitude. Let $LD$ be a perpendicular from $L$ on the side $BC$ . Determine the angles of $\triangle ABC$ in case if $AB \parallel HD$ .

Solution

Firstly, we consider the case when point $L$ belongs to the segment $AH$ (fig. 21). Since $\angle LHB = \angle LDB = 90^\circ$ , then $BLHD$ is cyclic quadrilateral. Thus, $\angle LBD = \angle DHC$ . On the other hand $\angle LBD = \angle LBA$ , due to the fact that $BL$ is a bisector and $\angle DHC = \angle BAC$ , because $AB \parallel HD$ . It follows that $\angle ABC = 2\angle BAC$ and $\angle ABC + \angle BAC = 180^\circ - 60^\circ = 120^\circ$ . And we get $\angle BAC = 40^\circ$ as $\angle ABC = 80^\circ$ .

We point out that $\angle LBD = \angle DHL$ and again $\angle LBD = \angle LBA$ , due to the fact that $BL$ is a bisector of the triangle and $\angle DHL = \angle BAC$ since $AB \parallel HD$ . Similarly we get $\angle BAC = 40^\circ$ and $\angle ABC = 80^\circ$ . But, if $\angle ACB = 60^\circ > \angle BAC = 40^\circ$ then the point $H$ can not belong to the segment $AL$ . Thus, this case is impossible.

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