Ex.10.6 Q7 Circles Solution - NCERT Maths Class 9

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\(\begin {align}{AC}\end {align}\) and \(\begin {align} {BD}\end {align}\) are chords of a circle which bisect each other. Prove that

(i) \(\begin {align} {AC}\end {align}\) and \(\begin {align} {BD}\end {align}\) are diameters,

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

 Video Solution
Ex 10.6 | Question 7

Text Solution

What is known?

\(\begin {align} {AC}\end {align}\) and \(\begin {align}{BD}\end {align}\) are \(2\) chords of a circle which bisect each other.

What is unknown?

Proof that \(\begin {align} {AC}\end {align}\) and \(\begin {align} {BD}\end {align}\) are diameters and \(\begin {align} {ABCD}\end {align}\) is a rectangle.


  • 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.
  • Using Side-Angle-Side (SAS criteria) and Corresponding parts of congruent triangles (CPCT) we prove the statement.


Let \(\begin {align} {AC}\end {align}\) and \(\begin {align} {BD}\end {align}\) be \(2\) chords intersecting at \(\begin {align} {O.}\end {align}\)

In \(\begin {align} \Delta {AOB}\end {align}\) and \(\begin {align} \Delta {COD,}\end {align}\)

\(\begin{align}{OA}&={OC} \quad (\text { Given }) \\ {OB}&={OD} \quad \text { (Given) } \\ \angle {AOB}&=\angle {COD} \\ \text {(Vertically } & \text{opposite}  \text{ angles }) \end{align}\)

\(∴ \Delta {AOB} \cong \Delta {COD}\) (SAS congruence rule) 

\( {AB} ={CD}\) (By CPCT)

Similarly, it can be proved that \(\begin{align} \Delta {AOD} \cong \Delta {COB} \end{align}\)

\(\begin{align} ∴ {AD}={CB} \qquad (\text{By CPCT}) \end{align}\)

Since in quadrilateral \(\begin{align} {ABCD,} \end{align}\) opposite sides are equal in length, \(\begin{align} {ABCD} \end{align}\) is a parallelogram.

We know that opposite angles of a parallelogram are equal.

\(\begin{align} ∴ \angle {A}=\angle {C} \end{align}\)


\(\begin{align} \angle {A} \!+ \! \angle {C} \!&=\!180^{\circ} \\{(ABCD} &\text { is a cyclic } \text{ quadrilateral) } \\\angle {A}+\angle {A} &=180^{\circ} \\ 2 \angle {A}&=180^{\circ} \\ ∴ \angle {A}&=90^{\circ}\end{align}\)

As \(\begin{align}{ABCD}\end{align}\) is a parallelogram and one of its interior angles is \(90^{\circ,}\) therefore, it is a rectangle.

\(\begin{align}\angle  {A}\end{align}\) is the angle subtended by chord \(\begin{align}\angle  {BD.}\end{align}\)

And as .\(\begin{align}\angle  {A} = 90^{\circ}, \end{align}\) therefore, \(\begin{align}  {BD}\end{align}\) should be the diameter of the circle. Similarly, \(\begin{align}  {AC}\end{align}\) is the diameter of the circle.

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