Question

Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA = QB

Answer 1

Let C (O, r) and C(O', r) be two equal circles. clearly, C(O, r) ≅ C(O', r).

Since PQ is a common chord of two congruent circles.

Therefore,

arc PCQ = arc PDQ

⇒ ∠QAP = ∠QBP

Two circles will be congruent if and only if they have equal radii

Here, PQ is the common chord to both circles.

Thus, their corresponding arcs are equal (if two chords of a circle are equal, then their corresponding arcs are congruent).

So, arc PCQ = arc PDQ

+ ∠QAP = ∠QBP (Congruent arcs have the same degree in measure)

Hence, QA = QB (In an isosceles triangle, base angles are equal)