Q15 of 47 Page 10

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.

Given: 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.
To prove: AB is the bisector of angle PBQ.

Proof:
∠ ABQ = ∠ QAX (Alternate interior angles)
∠ ABP = ∠ YAP (Alternate interior angles)
But ∠ YAP = ∠ XAQ (Vertically opposite angles)
∴ ∠ ABQ = ∠ ABP ∠ ABQ + ∠ ABP = 180o         (Linear pair)
∠ ABQ = ∠ ABP = 90o
AB is the bisector of angle PBQ.

More from this chapter

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13 Two rays ABP and ACQ are intersected by two parallel lines in B, C and P, Q respectively. Prove that the circumcircles of Δ ABC and Δ APQ touch each other at A. [Hint: Draw tangent XAY to the circumcircle of triangle APQ and show that ∠ YAP = ∠ PQA = ∠ BCA. 14 In figure, two circles touch internally at a point P. AB is a chord of the bigger circle touching the other circle at C. Prove that PC bisects the angle APB.[Hint: Draw a tangent at the point P. Joint CD, where D is the point of intersection of AP and the inner circle and prove that ∠ PBC = ∠ PCD.]
 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. 17 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)