Question

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

Answer 1

Given that $AD=BC$

Draw altitudes from $D$ and $C$ on $AB$ intersecting at $E$ and $F$ respectively.

In $△ADE$ and $△BCF$

$\Rightarrow \angle AED=\angle BCF\left({90}^0\right)$

$AD=BC$ (Given)

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

$\therefore △ADE\cong △BCF$ by R.H.S

$\therefore \angle DAE=\angle CBF$ (cpct) $\to \left(1\right)$

$\angle BAD+\angle ADC={180}^0$ \rightarrow(2)$

(Sum of cointesion angles $={180}^0$ $\because$ Here $DC\parallel AB$)

$\because \angle BAD$ or $\angle DAE=\angle CBF$ (From(1))

Putting in (2)

$\Rightarrow \angle CBF+angleADC={180}^0$

$\Rightarrow \angle B+\angle D={180}^0$

Hence, in $ABCD$ , sum of opposite angles $={180}^0$

$\therefore ABCD$ is a cyclic quadrilateral.

Hence, proved.