Question

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer 1

Step 1: Given Data

$ABCD$ is a trapezium $AB\parallel DC$ & non parallel sides are equal, i.e., $AB=BC$

Step 2: To prove

$ABCD$ is cyclic quadrilateral.

Step 3: Construction

Draw $DE\perp AB$ & $CF\perp AB$

Step 4: To Proof

To prove $ABCD$is cyclic quadrilateral, we prove that sum of one pair of opposite angle is ${180}^{\circ }$

In $△ADE\&△BCF$

$\angle AED=\angle BCF$ (Both ${90}^{\circ }$as $AE\perp DC$ & $BF\perp DC$)

$AD=BC$ (Given)

$DE=CF$ (Distance between parallel sides is equal)

$\therefore △ADE\cong △BCF$ (R-H-S congruence rule )

So,$\angle DAE=\angle BCF$ (By C.P.C.T.)

i.e.,$\angle A=\angle B....\left(1\right)$

Now, for parallel lines $AB$ and $DC$, $AD$is transversal line

$\angle A+\angle D={180}^{\circ }$ (Interior angles on the same side of transversal are supplementary)

$\Rightarrow \angle B+\angle D={180}^{\circ }$ (From equation $1$)

So, in $ABCD$ the sum of one pair of opposite angle is ${180}^{\circ }$

Therefore,$ABCD$ is a cyclic quadrilateral.

Hence proved.