# Ex.10.6 Q6 Circles Solution - NCERT Maths Class 9

## Question

\(\begin {align}{ABCD} \end {align}\) is a parallelogram. The circle through \(\begin {align}{A, B} \end {align}\) and \(\begin {align}{C} \end {align}\) intersect \(\begin {align}{CD} \end {align}\) (produced if necessary) at \(\begin {align}{E} \end {align}\). Prove that \(\begin {align}{AE = AD.} \end {align}\)

## Text Solution

**What is known?**

\(\begin {align}{ABCD} \end {align}\) is a parallelogram.

**What is unknown?**

Proof that \(\begin {align}{AE = AD} \end {align}\)

**Reasoning:**

- A quadrilateral \(\begin {align}{ABCD} \end {align}\) is called cyclic if all the four vertices of it lie on a circle.
- The sum of either pair of opposite angles of a cyclic quadrilateral is \(180^{\circ}.\)
- Opposite angles in a parallelogram are equal.

**Steps:**

We can see that \(\begin {align}{ABCE} \end {align}\) is a cyclic quadrilateral.

We know that in a cyclic quadrilateral, the sum of the opposite angles is \(180^{\circ.}\)

\[\begin{align}\angle {AEC}+\angle {CBA}&=180^{\circ} \\ \angle {AEC} +\angle {AED}&=180^{\circ} (\text { Linear pair })\end{align}\]

\(\angle {AED}=\angle {CBA} \ldots \text { (1) }\)

We know that in a parallelogram, opposite angles are equal.

\(\begin {align} \angle {ADE}=\angle{CBA}\quad \dots(2) \end {align}\)

From (**\(1\)**) and (**\(2\)**),

\[\begin{align} \angle {AED}=\angle{ADE}\\{AD}={AE}\qquad \end{align}\]

(sides opposite to equal Angles of a triangle are equal).