Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.4 National Olympiad Prove it Slovenia

Let $ABC$ be an acute triangle such that $|AB| > |BC| > |AC|$ . Let $D$ be a point different from $C$ on the segment $BC$ , such that $|AC| = |AD|$ . Let $H$ denote the orthocentre of the triangle $ABC$ , and let $A_1, B_1$ be the feet of the altitudes from $A$ and $B$ , respectively. Denote the intersection of the lines $A_1B_1$ and $DH$ by $E$ . Prove that $B, D, B_1$ and $E$ are concyclic.

Solution

Let $C_1$ be the foot of the altitude to the side $AB$ . The triangle $CAD$ is isosceles since $|AC| = |AD|$ .

The line $AA_1$ is the altitude in this isosceles triangle, so $\angle CDH = \angle HCD = \angle C_1CB = \frac{\pi}{2} - \angle CBA$ . This implies that
$\begin{aligned} \angle EDB &= \pi - \angle CDE = \pi - \angle CDH \\ &= \frac{\pi}{2} + \angle CBA. \end{aligned}$
In the quadrilateral $ABA_1B_1$ we have $\angle AA_1B = \frac{\pi}{2} = \angle AB_1B$ . So this is a cyclic quadrilateral and $\angle AB_1A_1 = \pi - \angle A_1BA = \pi - \angle CBA$ . We see that $\angle A_1B_1C = \pi - \angle AB_1A_1 = \angle CBA$ and
$\angle EB_1B = \angle EB_1A + \angle AB_1B = \angle A_1B_1C + \frac{\pi}{2} = \angle CBA + \frac{\pi}{2}.$
We have shown that $\angle EDB = \frac{\pi}{2} + \angle CBA = \angle EB_1B$ , which implies that $B, D, B_1$ and $E$ are concyclic.

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