  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. Mathematics

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