Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 5.0 AIME, harder Prove it Romania

Let $ABCD$ be a square and $E$ be a point on its diagonal $BD$ , different from its midpoint. Denote $H$ and $K$ the orthocenters of the triangles $ABE$ , respectively $ADE$ . Prove that $\overline{BH} + \overline{DK} = 0$ .

Mihaela Berindeanu

Solution

Notice that the points $H$ and $K$ are on the diagonal $AC$ , because $AC$ is perpendicular on $BE$ and $DE$ . Also, $H$ and $K$ are on the altitudes from $E$ in the two triangles, which are perpendicular on the sides of the initial square.

It follows that the triangle $EHK$ is right and isosceles, so $H$ and $K$ are symmetric with respect of the square's center. Since $B$ , $D$ are also symmetric with respect of the square's center, it follows that $DKBH$ is a parallelogram, whence the conclusion.

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