Q9:

In Fig. 10.13, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^{\circ}$.

Given:

1. A circle with center $O$.

2. Two parallel tangents $XY$ and $X'Y'$ touching the circle at points $P$ and $Q$ respectively.

3. A third tangent $AB$ touching the circle at point $C$, intersecting $XY$ at $A$ and $X'Y'$ at $B$.

To Prove:

$\angle AOB = 90^{\circ}$

Step 1: Construction and Identification of Congruent Triangles

Join $OC$. Consider $\triangle OPA$ and $\triangle OCA$.

In $\triangle OPA$ and $\triangle OCA$:

$OP = OC$ [Radii of the same circle]

$OA = OA$ [Common side]

$AP = AC$ [Tangents drawn from an external point $A$ to the circle are equal in length]

Therefore, $\triangle OPA \cong \triangle OCA$ [By SSS Congruence Criterion].

Consequently, $\angle POA = \angle COA$ (Equation 1) [By CPCT - Corresponding Parts of Congruent Triangles].

Step 2: Applying Congruence to the Second Set of Triangles

Similarly, consider $\triangle OQB$ and $\triangle OCB$.

In $\triangle OQB$ and $\triangle OCB$:

$OQ = OC$ [Radii of the same circle]

$OB = OB$ [Common side]

$BQ = BC$ [Tangents drawn from an external point $B$ to the circle are equal in length]

Therefore, $\triangle OQB \cong \triangle OCB$ [By SSS Congruence Criterion].

Consequently, $\angle QOB = \angle COB$ (Equation 2) [By CPCT].

Step 3: Summing the Angles on the Straight Line

$PQ$ is a diameter of the circle because $XY \parallel X'Y'$ and the tangents are perpendicular to the diameter at the points of contact. Thus, $POQ$ is a straight line.

The sum of all angles formed at the center $O$ on the straight line $PQ$ is $180^{\circ}$:

$\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}$

Step 4: Substitution and Final Calculation

Using Equation 1 ($\angle POA = \angle COA$) and Equation 2 ($\angle QOB = \angle COB$):

$\angle COA + \angle COA + \angle COB + \angle COB = 180^{\circ}$

$2\angle COA + 2\angle COB = 180^{\circ}$

$2(\angle COA + \angle COB) = 180^{\circ}$

$\angle COA + \angle COB = \frac{180^{\circ}}{2}$

$\angle AOB = 90^{\circ}$ [Since $\angle COA + \angle COB = \angle AOB$]

Final Answer: $\angle AOB = 90^{\circ}$