Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National Olympiad Prove it Ireland

On the sides $AB$ and $AC$ of triangle $ABC$ the triangles $AEB$ and $ADC$ are constructed, both similar to $\triangle ABC$ and with $\angle AEB = \angle ADC = \angle BAC$ . Prove that the area of $\triangle ABC$ is less than, equal to or greater than the sum of the areas of triangles $AEB$ and $ADC$ according as $\angle BAC$ is less than, equal to or greater than a right angle.

Solution

W.l.o.g we can arrange that $\angle ABE = \angle ABC$ and $\angle ACD = \angle ACB$ . Let $D'$ and $E'$ be the reflections of $D$ and $E$ in $AC$ and $AB$ , respectively. Then $D'$ and $E'$ are on $BC$ , $\angle D'AC = \angle DAC = \angle ABC$ and $\angle E'AB = \angle EAB = \angle BCA$ .

If $\angle BAC > 90^\circ$ then $\angle D'AC + \angle E'AB = \angle ABC + \angle BCA < 90^\circ < \angle BAC$ .
In this case the triangles $AE'B$ and $AD'C$ do not overlap and so the sum of their areas is less than the area of triangle $ABC$ .
If $\angle BAC < 90^\circ$ then $\angle D'AC + \angle E'AB = \angle ABC + \angle BCA > 90^\circ > \angle BAC$ .
In this case the triangles $AE'B$ and $AD'C$ overlap and so the sum of their areas is greater than the area of triangle $ABC$ .
If $\angle BAC = 90^\circ$ then $\angle D'AC + \angle E'AB = \angle ABC + \angle BCA = 90^\circ = \angle BAC$ .
In this case the triangles $AE'B$ and $AD'C$ exactly cover triangle $ABC$ and so the sum of their areas is equal to the area of triangle $ABC$ .

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