# Ex.10.5 Q8 Circles Solution - NCERT Maths Class 9

## Question

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

## Text Solution

**What is known?**

Non-parallel sides of trapezium are equal.

**What is unknown?**

To prove that trapezium is cyclic.

**Reasoning:**

If the sum of a pair of opposite angles of a quadrilateral is \(180^{\circ},\) the quadrilateral iscyclic. Using Right angled-Hypotenuse-Side (**RHS**) criteria and Corresponding parts of congruent triangles (**CPCT**) we prove the statement.

**Steps:**

Draw a trapezium \( {ABCD} \)

with \( {AB} \| {CD} \)

\(AD\) and \(BC\) are the non-parallel sides which are equal. \({AD = BC.}\)

Draw \(\begin {align} {AM} \perp {CD} \end {align}\)

and \(\begin {align} {BN} \perp {CD} \end {align}\)

Consider \(\Delta {AMD}\) and \(\Delta \text{BNC.}\)

\(\begin{align} {AD} &={BC} \qquad \; (\text { Given }) \\\angle{AMD}&=\angle {BNC} \quad \left(90^{\circ}\right) \end{align}\)

\(AM =BN \) (Perpendicular distance between two parallel lines is same)

By **RHS** congruence,

\[\begin {align}\Delta {AMD} \cong \Delta {BNC} \end {align}\]

Using CPCT ,

\[\begin {align} \angle {ADC}=\angle {BCD} \ldots . .(1) \end {align}\]

\(\angle {BAD} \) and \(\angle {ADC} \) are on the same side of transversal \(\angle \text{AD}. \)

\[\begin{align}\angle {BAD}+\angle {ADC}&=180^{\circ} \\ \angle {BAD}+\angle {BCD}&=180^{\circ} \text{[Using}(1)]\end{align}\]

This equation proves that the opposite angles are supplementary.

Hence, \({ABCD} \) is a cyclic quadrilateral.