Question

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 ∠AOB = 90°.

Fig. 10.13

Answer 1

Let us join point O to C.

In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
ΔOPA ≅ ΔOCA (SSS congruence criterion)
Therefore, P ↔ C, A ↔ A, O ↔ O
∠POA = ∠COA         ....… (i)
Similarly, ΔOQB ≅ ΔOCB
∠QOB = ∠COB         ....… (ii)
Since POQ is a diameter of the circle, it is a straight line.
Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii), it can be observed that
2∠COA + 2 ∠COB = 180 º
∠COA + ∠COB = 90 º
∠AOB = 90°

Answer 2

We joined points O and C.

According to the question, in$\Delta OPA$ and $\Delta OCA:$

• $OP=OC$ (Radii of the same circle)
• $AP=AC$ (Tangents from point A)
• $AO=AO$ (Common side)

Therefore, by the S congruence criterion: $\Delta OPA \cong \Delta OCA$

This implies: $\angle POA = \angle COA \quad \text{(i)}$

Similarly, in $\Delta OQB$ and $\Delta OCB:$

• $OQ=OC$ (Radii of the same circle)
• $BQ=BC$ (Tangents from point B)
• $BO=BO$ (Common side)

Therefore, by the S congruence criterion: $\Delta OQB \cong \Delta OCB$

This implies: $\angle QOB = \angle COB \quad \text{(ii)}$

Since POQ is a diameter of the circle, it is a straight line: $\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$

From equations (i) and (ii): $2\angle COA + 2\angle COB = 180^\circ$

Dividing by 2: $\angle COA + \angle COB = 90^\circ$

Thus:$\angle AOB = 90^\circ$