Competition problem — Geometry — Maths Olympiad Prep

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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. Fractions can be typed as 3/2, and spacing doesn't matter.

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 $\angle ABC = 180^{\circ} - \angle BAC - \angle BCA = 180^{\circ} - 70^{\circ} - 70^{\circ} = 40^{\circ}$ .

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