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:\( \angle AOB = 90^\circ\)