Question

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

• A. EA = EB
• B. EA = AD
• C. AD = BC
• D. BC = EB

Answer 1

Answer: (A), (C)

The correct options are
A EA = EB
C AD = BC

In $△$EDC and $△$EBA, we have

$\angle$1 = $\angle$2
$\angle$3 = $\angle$4
($\because$ alternate angles are equal)

and $\angle$CED = $\angle$AEB
($\because$ vertically opposite angles are equal)

$\therefore △EDC\sim △$EBA (By AA similarity criterion)

$\Rightarrow \frac{ED}{EB}$ = $\frac{EC}{EA}$

$\Rightarrow \frac{ED}{EC}$ = $\frac{EB}{EA}$ ......(i)

It is given that $△$AED $\sim$ $△$BEC

$\therefore \frac{ED}{EC}$ = $\frac{EA}{EB}$ = $\frac{AD}{BC}$ ......(ii)

From (i) and (ii), we get

$\frac{EB}{EA}$ = $\frac{EA}{EB}$

$\Rightarrow (EB{)}^2$ = $(EA{)}^2$

$\Rightarrow EB=EA$

On substituting EB = EA in (ii), we get

$\frac{EA}{EA}$ =$\frac{AD}{BC}$

$\Rightarrow \frac{AD}{BC}$ = 1

$\therefore AD=BC$