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°`
In quadrilateral OAPB
`∠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
AD || BC
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.
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