In figure, ABCD is a trapezium with AB||DC, If $△$ AED is similar to $△$ BEC, prove that AD = BC

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Solution

In $△$ EDC and $△$ EBA, we have

$∠$ 1 = $∠$ 2 [alternate angles]

$∠$ 3 = $∠$ 4 [alternate angles]

and, $∠$ CED = $∠$ AEB [vertically opposite angles]

$∴ △ E D C$ ~ $△$ EBA

$\Rightarrow \frac{E D}{E B}$ = $\frac{E C}{E A}$

$\Rightarrow \frac{E D}{E B}$ = $\frac{E C}{E A}$ ......(i)

It is given that $△$ AED ~ $△$ BEC

$∴ \frac{E D}{E C}$ = $\frac{E A}{E B}$ = $\frac{A D}{B C}$

From (i) and (ii), we get

$\frac{E B}{E A}$ = $\frac{E A}{E B}$

$\Rightarrow (E B)^{2}$ = $(E A)^{2}$

$\Rightarrow E B = E A$

Substituting EB = EA in (ii), we get

$\frac{E A}{E A}$ = $\frac{A D}{B C}$

$\Rightarrow \frac{A D}{B C}$ = 1

$∴ A D = B C$

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In figure, ABCD is a trapezium with AB||DC. If $△$ AED is similar to $△$ BEC, which of the following is true?

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