Competition problem — Geometry — Maths Olympiad Prep

Library / /127 of 144

Geometry Difficulty 2.6 Junior Find the answer

In $\triangle ABC$ , points $E$ and $F$ are on $AB$ and $BC$ , respectively, such that $AE = BF$ and $BE = CF$ . If $\angle BAC = 70^{\circ}$ , what is the measure of $\angle ABC$ ?

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Since $AE = BF$ and $BE = CF$ , then $AB = AE + BE = BF + CF = BC$ . Therefore, $\triangle ABC$ is isosceles with $\angle BAC = \angle BCA = 70^{\circ}$ . Since the sum of the angles in $\triangle ABC$ is $180^{\circ}$ , then $\angle ABC = 180^{\circ} - \angle BAC - \angle BCA = 180^{\circ} - 70^{\circ} - 70^{\circ} = 40^{\circ}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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