Q7 of 12 Page 164

In the figure, O is the centre of a circle. Tangents PA and PB drawn from an external point P touch the circle at points A and B respectively. Prove that OP is perpendicular bisector of AB.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.


∠ OAP = ∠ OBP = 90°


Now, ∠ OAP + ∠ APB + ∠ OBP + ∠ AOB = 360° [Angle sum property of quadrilaterals]


⟹ 90° + ∠ APB + 90° + ∠ AOB = 360°


⟹ ∠ AOB = 360° - 180° - ∠ APB = 180° - ∠ APB.... (1)


Now, in Δ OAB, OA is equal to OB as both are radii.


⟹ ∠ OAB = ∠ OBA [In a triangle, angles opposite to equal sides are equal]


Now, on applying angle sum property of triangles in Δ AOB,


We obtain ∠ OAB + ∠ OBA + ∠ AOB = 180°


⟹ 2∠ OAB + ∠ AOB = 180°


⟹ 2 ∠ OAB + (180° - ∠ APB) = 180° [Using (1)]


⟹ 2 ∠ OAB = ∠ APB


Thus, the given result is proved


More from this chapter

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5

A circle with centre O touches all the four sides of a quadrilateral ABCD. If the point of contact divides AB in the ratio 3 : 1 and AB = 8 cm, then find the radius of the circle if it is given OA = 10 cm.

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A circle touches the sides of a quadrilateral. Show that the angles subtended at the centre by a pair of opposite sides are supplementary.

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In the figure, O is the centre of the circle. Tangents PA and PB drawn from an external point P touch the circle at A and B respectively. Prove that PAOB is a cyclic quadrilateral.

 1

Answer the following questions from the figure.


(i) Alternate segment of ∠BAQ.


(ii) Alternate segment of ∠DAP.


(iii) If C is jointed to B, then ∠ACB is equal to which angles?


(iv) ∠ABD and ∠ADB are equal to which angles?