Question

In the given figure $∠ A D E = ∠ A CB$ and $\frac{A D}{D B} = \frac{A E}{EC}$ . Prove that $△ A BC$ is an isosceles triangle.

Approach

Since $\frac{A D}{D B} = \frac{A E}{EC}$ , by the converse of Basic Proportionality Theorem, $D E ∥ BC$ . Thus $∠ A D E = ∠ A BC$ (corresponding angles). Given $∠ A D E = ∠ A CB$ , we have $∠ A BC = ∠ A CB$ , so $A C = A B$ , making $△ A BC$ isosceles.

Step-by-step working

1
$\frac{A D}{D B} = \frac{A E}{EC} \Rightarrow D E ∥ BC$ (converse of BPT).
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2
$∠ A D E = ∠ A BC$ (corresponding angles).
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3
Given $∠ A D E = ∠ A CB$ , so $∠ A BC = ∠ A CB$ .
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4
$∠ A BC = ∠ A CB \Rightarrow A C = A B$ , hence $△ A BC$ is isosceles.
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Concepts used

converse of bptisosceles trianglesimilar triangles

Approach

Step-by-step working

Concepts used

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