Q17 of 47 Page 10

If PAB is a secant to a circle intersecting it at A and B, and PT is a tangent, then PA. PB = PT2.
(Prove using alternate segment theorem)

Given: A secant PAB to circle with centre O intersecting it at A and B and a tangent PT to the circle.

To prove: PA. PB = PT2
Construction: Join TA and TB.
Proof:
In Δ PBT and Δ PTA
∠ BPT = ∠ APT (Common)
∠ PBT = ∠ PTA (Alternative segment theorem)
∴ Δ PBT ∼ Δ PTA (AA similarity)

PA.PB = PT2
Hence proved.

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15 Two circles intersect each other at two points A and B. At A, tangents AP and AQ to the two circles are drawn which intersect other circles at the points P and Q respectively. Prove that AB is the bisector of angle PBQ. 16 The diagonals of a parallelogram ABCD intersect in a point E. Show that the circumcircles of Δ ADE and Δ BCE touch each other at E. 18 A circle touches the side BC of a triangle ABC at P and touches AB and AC when produced at Q and R respectively. Show that AQ = ½ (perimeter of triangle ABC). 19 Prove that the intercept of a tangent between 2 parallel tangents to a circle subtends a right angle at the centre.