Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Slovenia

The diagonal $BD$ of the quadrilateral $ABCD$ divides this quadrilateral into an acute triangle $ABD$ and an equilateral triangle $BCD$ . Let $O$ be the orthocentre of the triangle $ABD$ . Prove:

a. if the triangles $ABD$ and $OCD$ are congruent, then $AB \perp BC$ ;

b. if $\angle CBA = 90^\circ$ , then the triangles $ABD$ and $OCD$ are congruent.

Solution

a. The points $C$ and $O$ lie on the bisector of the segment $BD$ , so $\angle DCO = 30^\circ$ . Since the triangles $ABD$ and $OCD$ are congruent, we have $\angle DBA = \angle DCO = 30^\circ$ . This implies $\angle CBA = \angle CBD + \angle DBA = 90^\circ$ .

b. If $\angle ABC = 90^\circ$ , then $\angle ABD = 30^\circ$ . The points $C$ and $O$ lie on the bisector of the segment $BD$ . So $\angle DCO = 30^\circ$ . Applying the formula connecting the inscribed and the central angle we see that $\angle BOD = 2\angle BAD$ . On the other hand $\angle BOD = 2\angle COD$ , so $\angle COD = \angle BAD$ . The triangles $ABD$ and $OCD$ are congruent since $\angle BAD = \angle COD$ , $\angle DBA = \angle DCO$ and $|BD| = |CD|$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.