In given figure, ABCD is a trapezium with AB||DC. If $△ A E D$ is similar to $△ B E C$ . prove that AD=BC.

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Solution

In $△ E D C$ and $△ E B A$ , we have

$∠ 1 = ∠ 2$ [Alternate angles]

$∠ 3 = ∠ 4$ [Alternate angles] and,

$∠ C E D = ∠ A E B$ [Vertically opposite angles]

$∴$ $△ E D C \sim △ E B A$

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

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

It is given that $△ A E D \sim △ B E C$

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

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

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

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

$\Rightarrow$ $A D = B C$ $[H e n c e p r o v e d]$

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Similar questions

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

A B C D, A B ∥ D C

△ A E D \sim △ B E C

A D = B C .

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

A B C D

A B

D C

A D = B C

O A = O B

O C = O D

□

\frac{1}{2}

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