Question

In the following figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.

Answer 1

Given:

O is the centre of the circle.

The line B–C–D is tangent to the circle at C.

A, O and the point opposite A on the circle are collinear (so AO is along a diameter / BA is the diameter produced) — as shown in the figure.

To Prove: ∠BAC + ∠ACD = 90°.

Proof [Step-wise]:

1. OA and OC are radii of the circle, so OA = OC.

Hence triangle AOC is isosceles and its base angles are equal: ∠OAC = ∠ACO.

2. The radius drawn to the point of contact is perpendicular to the tangent.

Since BCD is tangent at C, OC ⟂ CD.

3. Because OC ⟂ CD, the angle made at C by AC and the tangent CD is complementary to the angle between AC and OC. In other words, ∠ACO + ∠ACD = 90°. (OC and CD form a right angle, so the two angles from AC to OC and from AC to CD add to 90°.)

4. From (1) we have ∠OAC = ∠ACO.

From the figure (A, O, B are collinear), ∠BAC = ∠OAC. 

Therefore ∠BAC = ∠ACO.

5. Replace ∠ACO in step (3) by ∠BAC (using step 4): ∠BAC + ∠ACD = 90°.

Thus ∠BAC + ∠ACD = 90°, as required.