Circle

Exercise 10.2 Part 5

Question: 10 - Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Answer: Construction: Draw a circle with centre O. Tangent PA and PB are drawn to circle.

To Prove: ∠APB + ∠AOB = 180°

∠OAP = ∠OBP = 90°

∠AOB + ∠APB + ∠OAP + ∠OBP = 360° (sum of all angles of a quadrilateral)

Or, ∠AOB + ∠APB + 90° + 90° = 360°

Or, ∠AOB + ∠APB = 360° - 180° = 180° proved

Question: 11 - Prove that the parallelogram circumscribing a circle is a rhombus.

Answer: Construction: Draw a circle with centre O. Draw a parallelogram ABCD which touches the circle at P, Q, R and S.

Given; AB || DC

To Prove: ABCD is a rhombus

In ΔAOB and ΔDOC;

AB = DC (given that AD || BC)

∠AOB = ∠DOC (Vertically opposite angles)

∠BAO = ∠DCO (Alternate angles)

Hence; ΔAOB ≅ ΔDOC

Hence; AO = CO and BO = DO

Since diagonals are bisecting each other, so given parallelogram is a rhombus.