Question

The angles subtended by the intercept of a tangent between two parallel tangents to the circle is

Answer 1

Answer on Question #48835 – Math – Geometry

Question:

The angles subtended by the intercept of a tangent between two parallel tangents to the circle is ...

Answer:

The angles subtended by the intercept of a tangent between two parallel tangents to the circle is $90{}^{\circ}$.

Given: XY and X'Y' at are two parallel tangents to the circle with centre O and AB is the tangent at the point C, which intersects XY at A and X'Y' at B.

To prove: $\angle AOB = 90{}^\circ$.

Construction: Join OC.

Proof:

In $\Delta OPA$ and $\Delta OCA$, we have

$OP = OC$ (Radii of the same circle)

$AP = AC$ (Tangents from point A)

$AO = AO$ (Common side)

$\Delta OPA \cong \Delta OCA$ (SSS congruence criterion)

Therefore, $\angle POA = \angle COA \quad \ldots (1)$ (C.P.C.T)

Similarly, $\Delta OQB \cong \Delta OCB$

$\angle QOB = \angle COB \quad \ldots (2)$

POQ is a diameter of the circle. Hence, it is a straight line.

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

From equations (1) and (2), it can be observed that

$$2 \angle COA + 2 \angle COB = 180{}^\circ$$ $$\therefore \angle COA + \angle COB = 90{}^\circ$$ $$\therefore \angle AOB = 90{}^\circ$$

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