Question

If two non-parallel sides of a trapezium are equal prove that it is cyclic

Answer 1

Solve to prove that if two non-parallel sides of a trapezium are equal prove that it is cyclic

Given:

$ABCD$ is a trapezium where non-parallel sides are equal that is, $AD=BC$

Construction:

$DM$ and$CN$are perpendicular that is drawn on $AB$from$D$and$C$respectively.

To prove:

$ABCD$is cyclic trapezium

Proof:

In $\Delta DAM$ and $\Delta CBN,$

$AD=BC$

$\angle DAM=\angle CBN$(Non-parallel sides are equal then bases angles will be equal)

$\angle AMD=\angle BNC$ (Both right angles)

$\Rightarrow △DAM\cong △CBN$(By A.SA. property)

$DM=CN$ (By CPCT)

Therefore,

$\Delta DAM\cong \Delta CBN$ by RHS congruence condition

$$\begin{gathered} \angle A=\angle B \\ \angle B+\angle C=180° \\ \angle A+\angle C=180° \end{gathered}$$ ;($\angle CDM=\angle DMN={90}^{\circ }$)

Hence, $ABCD$is a cyclic quadrilateral as the sum of the pair of opposite angles is $180°$