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Question

In the trapezium $$ABCD,\; AB \parallel DC$$ and $$\triangle AED \sim \triangle BEC$$. Prove that $$AD = BC.$$


Solution

To Prove: $$AD = BC$$

Proof: Compare $$\triangle EDC$$ and $$\triangle EBA$$

$$\angle EDC = \angle EBA$$             [Alternate Angles]

$$\angle ECD = \angle EAB$$             [Alternate Angles]

By $$AA$$ criterion of similarity,
$$\triangle EDC \sim \triangle EBA$$

$$\Rightarrow \dfrac{ED}{EB} = \dfrac{EC}{EA}$$          [By CPST]

$$\Rightarrow \dfrac{ED}{EC} = \dfrac{EB}{EA}\quad\quad\quad\dots(i)$$

But, $$\triangle AED \sim \triangle BEC$$                 [Given]

$$\Rightarrow \dfrac{ED}{EC} = \dfrac{EA}{EB} = \dfrac{AD}{BC}$$            $$\dots(ii)$$ [By CPST]

$$\dfrac{EB}{EA} = \dfrac{EA}{EB} $$                         [From $$(i)$$ and $$(ii)$$]

$$\Rightarrow EA^2 = EB^2$$

$$\Rightarrow EA = EB$$

Hence, from equation $$(ii),$$
$$\dfrac{AD}{BC} = 1$$ 
$$\Rightarrow AD = BC$$

Hence proved.

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Mathematics

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