In the given figure, $\angle D = \angle E$ and $\frac{AD}{DB} = \frac{AE}{EC}$. Prove that $BAC$ is an isosceles triangle.

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Question

In the given figure, $\angle D = \angle E$ and $\frac{AD}{DB} = \frac{AE}{EC}$. Prove that $BAC$ is an isosceles triangle.

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Accepted Answer

Step 1: $\angle D = \angle E \Rightarrow AE = AD$
Step 2: $\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow DB = EC$
Step 3: $	herefore AD + DB = AE + EC \Rightarrow AB = AC$
Hence $	riangle BAC$ is an isosceles triangle.

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