Track / Stage 6 / 189 of 400 #1189 of 1964

Problem 1189

National olympiad, first round

Geometry Difficulty 6.2 Prove it

4. In a quadrilateral $A B C D$ the intersection of the diagonals is called $P$ . Point $X$ is the orthocentre of triangle $P A B$ . (The orthocentre of a triangle is the point where the three altitudes of the triangle intersect.) Point $Y$ is the orthocentre of triangle $P C D$ . Suppose that $X$ lies inside triangle $P A B$ and $Y$ lies inside triangle $P C D$ . Moreover, suppose that $P$ is the midpoint of line segment $X Y$ .

Prove that $A B C D$ is a parallelogram.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

4. [Version klas 5 \& klas 4 and below] Let $K$ be the intersection of $A X$ and $B D$ , and let $L$ be the intersection of $C Y$ and $B D$ . Consider the triangles $P L Y$ and $P K X$ . The angles $\angle P L Y$ and $\angle P K X$ are both right angles. The angles $\angle Y P L$ and $\angle X P K$ are opposite angles and therefore equal. Since $|P X|=|P Y|$ , we see that triangles $P K X$ and $P L Y$ are congruent (SAA). Hence, $|P K|=|P L|$ .

Now consider triangles $P A K$ and $P C L$ . Angles $\angle A K P$ and $\angle C L P$ are both right angles. Angles $\angle K P A$ and $\angle L P C$ are opposite angles, hence equal. We have already shown that $|P K|=|P L|$ .

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Therefore, triangles $P A K$ and $P C L$ are congruent (ASA). From this, we conclude that $|A P|=$ $|P C|$ .

In a similar fashion, we may deduce that $|B P|=|D P|$ . The two diagonals of $A B C D$ bisect each other, hence $A B C D$ is a parallelogram.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.