# Ex.10.2 Q9 Circles Solution - NCERT Maths Class 10

## Question

In Figure, \(XY\) and \(X'Y' \) are two parallel tangents to a circle with center \(O\) and another tangent \(AB\) with point of contact \(C\) intersecting \(XY\) at \(A\) and \(X'Y' \) at \(B.\)

Prove that \(\angle {AOB} = {90^ \circ }\).

## Text Solution

**What is known?**

- '\(O\)' is the centre of the circle.
- \(XY\) and \(X'Y'\)' are the two parallel tangents to the circle.
- \(AB\) is another tangent with point of contact \(C\), intersecting \(XY\) at \(A\) and \(X'Y'\) at \(B.\)

**To prove:**

\(\angle {AOB} = {90^ \circ }\)

**Reasoning :**

Join point \(O\) to \(C.\)

In \(\Delta OPA\) and \(\Delta OCA\)

\(OP = OC\) (Radii of the circle are equal)

\(AP = AC \) (The lengths of tangents drawn from an external point \(A\) to a circle are equal.)

\(AO = AO\) (Common)

By **\(SSS\)** congruency, \(\Delta {OPA} \cong \Delta {OCA}\)

**SSS Congruence Rule:** If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

**Steps:**

Therefore, \(\angle {OPA} = \angle {OCA}\)

Similarly, \(\Delta {OCB} \cong \Delta {OBQ}\)

Therefore, \(\angle {COB} = \angle {BOQ}\)

\(PQ\) is a diameter, hence a straight line and \(\angle {POQ} = {180^ \circ}\)

But \(\begin{align} \angle {P O Q} =\begin{bmatrix} \angle {P O A} + \angle {A O C}+ \\ \angle {C O B} + \angle {B O Q}\end{bmatrix}\end{align}\)

Therefore,

\[\begin{align} & \left[ \begin{array} \angle POA+\angle AOC+ \\ \angle COB+\angle BOQ \\ \end{array} \right]={{180}^{{}^\circ }} \\ & 2\angle AOC+ 2\angle COB ={{180}^{{}^\circ }}\end{align}\]

Therefore,

\[\begin{align}& \left[ \begin{array}& \angle POA=\angle AOC \;\text{and} \\\angle COB=\angle BOQ \\ \end{array} \right] \end{align}\]

Therefore,

\[\angle AOC+\angle COB={{90}^{{}^\circ }}\]

From the figure, \(\angle {A O C} + \angle {C O B} = \angle {A O B}\)

\(\therefore \quad \angle {A O B} = 90 ^ { \circ }\)

Hence Proved