Question

In the adjoining figure, $\frac{A D}{B D} = \frac{A E}{EC}$ and $∠ B D E = ∠ CE D$ , prove that $△ A BC$ is an isosceles triangle.

Approach

In triangle ABC with $\frac{A D}{D B} = \frac{A E}{EC}$ and $∠ B D E = ∠ CE D$ , by the converse of the Basic Proportionality Theorem, $D E ∥ BC$ . Hence corresponding angles $∠ D BC = ∠ BCE$ , implying triangle ABC is isosceles with $A B = A C$ .

Step-by-step working

1
Given $\frac{A D}{D B} = \frac{A E}{EC}$ , by converse of BPT, $D E ∥ BC$ .
1 mark
2
Since $D E ∥ BC$ and $∠ B D E = ∠ CE D$ , corresponding angles give $∠ D BC = ∠ BCE$ , so $A B = A C$ (triangle ABC is isosceles).
1 mark

Concepts used

trianglesisosceles triangleprovesimilar triangles

Approach

Step-by-step working

Concepts used

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