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Question

In Figure, $$ABCD$$ is a trapezium with $$A B \parallel DC$$. If $$\triangle A E D$$ is similar to $$\Delta B E C,$$ prove that $$A D=B C$$


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Solution

In $$\Delta E D C$$ and $$\Delta E B A$$, we have 

$$\angle 1=\angle 2 \quad $$ [Alternate angles ]
$$\angle 3=\angle 4\quad $$ [Alternate angles]  and 
$$\angle C E D=\angle A E B \quad $$ [ Vertically opposite angles ]

$$\therefore  \Delta E D C \sim \Delta E B A$$    [By AA criterion of similarity]

$$\Rightarrow \dfrac{E D}{E B}=\dfrac{E C}{E A} \\ \Rightarrow \dfrac{E D}{EC}=\dfrac{E B}{E A} \quad \quad \dots (i)$$

It is given that $$\Delta {AED} \sim \Delta {BEC}$$

$$\therefore \dfrac{E D}{E C}=\dfrac{E A}{E B}=\dfrac{A D}{B C} \quad \quad\dots (ii)$$

From $$(i)$$ and $$(ii)$$, we get,
$$\dfrac{E B}{E A}=\dfrac{E A}{E B} \\ \Rightarrow (E B)^{2}=(E A)^{2} \\\Rightarrow E B=E A$$

Substituting $$EB = EA$$ in $$(ii)$$, we get

$$\dfrac{E A}{E A}=\dfrac{A D}{B C} \\ \Rightarrow  \dfrac{A D}{B C}=1\\ \Rightarrow  A D=B C$$

Mathematics

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