Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.2 Junior Find the answer

In $\triangle ABC$ , points $D$ and $E$ lie on $AB$ , as shown. If $AD=DE=EB=CD=CE$ , what is the measure of $\angle ABC$ ?

Your answer

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

Solution

Since $CD=DE=EC$ , then $\triangle CDE$ is equilateral, which means that $\angle DEC=60^{\circ}$ . Since $\angle DEB$ is a straight angle, then $\angle CEB=180^{\circ}-\angle DEC=180^{\circ}-60^{\circ}=120^{\circ}$ . Since $CE=EB$ , then $\triangle CEB$ is isosceles with $\angle ECB=\angle EBC$ . Since $\angle ECB+\angle CEB+\angle EBC=180^{\circ}$ , then $2 \times \angle EBC+120^{\circ}=180^{\circ}$ , which means that $2 \times \angle EBC=60^{\circ}$ or $\angle EBC=30^{\circ}$ . Therefore, $\angle ABC=\angle EBC=30^{\circ}$ .

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