Question

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

Answer 1

Step 1:- Given

1. According to the given details $ABCD$ is a trapezium in which $AD\left|\right|BC$
2. Non-parallel sides $ABandDC$ of the trapezium $ABCD$ are equal i.e., $AB=DC$.

Step 2:- Construction and congruent triangle:

If the sum of a pair of opposite angles of a quadrilateral is $180°$, the quadrilateral is cyclic.

Draw a trapezium $ABCD$ with $AD\left|\right|BC$

$\text{AD and BC}$ are the non-parallel sides that are equal. $AB=DC$.

Draw $\text{AM and DN}$ such that they are perpendicular to $BC.$

From, the right triangles $\mathrm{AMB}\mathrm{and}\mathrm{DNC},$

$$\begin{gathered} \angle AMB=\angle DNC=90° \\ AB=DC\left(Given\right) \end{gathered}$$

Since the perpendicular distance between two parallel lines are the same,

$$\begin{gathered} AM=DN \\ △AMB\cong △DNC\mathbf{}\mathbf{}\mathbf{(}\mathbf{By}\mathbf{}\mathbf{RHS}\mathbf{}\mathbf{congruence}\mathbf{}\mathbf{rule}\mathbf{)} \\ \angle B=\angle C\mathbf{\left(}\mathbf{CPCT}\mathbf{\right)} \\ and,\angle 1=\angle 2 \\ \angle BAD=\angle 1+90° \\ =\angle 2+90° \\ =\angle CDA \end{gathered}$$

From quadrilateral $ABCD$

$$\begin{gathered} \angle B+\angle C+\angle CDA+\angle BAD=360° \\ \Rightarrow \angle B+\angle B+\angle CDA+\angle CDA=360°\mathbf{[}\mathbf{\because }\mathbf{\angle }\mathbf{B}\mathbf{=}\mathbf{\angle }\mathbf{C}\mathbf{,}\mathbf{}\mathbf{\angle }\mathbf{BAD}\mathbf{=}\mathbf{\angle }\mathbf{CDA}\mathbf{]} \\ \Rightarrow 2(\angle B+\angle CDA)=360° \\ \Rightarrow \angle B+\angle CDA=180° \end{gathered}$$

We know that, if any pair of opposite angles of a quadrilateral is $180°$, then the quadrilateral is cyclic.

Hence, the trapezium $ABCD$ is cyclic.