Question

Prove that an isosceles trapezium is always cyclic.

Answer 1

Given : A trapezium ABCD in which $AB\parallel DC$ and $AD=BC$

To prove: ABCD is a cyclic quadrilateral

Construction : Draw $DE\perp AB,CF\perp AB$

Proof:

In $\Delta DEA$ and $\Delta CFB$, we have

(i) $AD=BC$ (Given)

(ii) $\angle DEA=\angle CFB={90}^{\circ }$ $(DE\perp ABandCF\perp AB)$

(iii) $DE=CF$ (Distance between parallel lines remains constant)

$\therefore$ $\Delta DEA\cong \Delta CFB$ (By RHS criteria)

$\angle A=\angle B$ ......(i) By CPCT

$\angle ADE=\angle BCF$ By CPCT

$\angle ADE+\angle EDC=\angle BCF+\angle DCF$

$(\because \angle EDC=\angle DCF={90}^{\circ })$

$\angle D=\angle C$......(ii)

$\angle ADE+\angle CDE=\angle D$,

$\angle BCF+\angle DCF=\angle C$

$\therefore \angle A=\angle B$ and $\angle C=\angle D$ .....from (i) and (ii)

$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$ (Since sum of angles of a quadilateral is ${360}^{\circ }$ )

2 $(\angle B+\angle D)={360}^{\circ }$ Using (i) and (ii)

Sum of a pair of opposite angles of a quadrilateral ABCD is ${180}^{\circ }$

Hence, it is proved that ABCD is a cyclic quadrilateral.