# 2007 BMO Problems/Problem 1

## Contents

## Problem

(*Albania*)
Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .

## Solution

Since , , and similarly, . Since , by considering triangles we have . It follows that .

Now, by the Law of Sines,

.

It follows that if and only if

.

Since ,

and

From these inequalities, we see that if and only if (i.e., ) or (i.e., ). But if , then triangles are congruent and , a contradiction. Thus we conclude that if and only if , Q.E.D.

## Solution 2

Let and . Then by the isosceles triangles manifest in the figure we have and , so and . Furthermore and .

If , then . But also , so by SSA "Incongruence" (aka. the Law of Sines: ) we have . This translates into , or , which incidentally equals , as desired.

If , then also by the Exterior Angle Theorem, so and hence and are supplementary. A simple Law of Sines calculation then gives , as desired. This completes both directions of the proof.

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*