Question

Timir drew two circles which intersect eachother at the points P and Q. Through the point P two straight lines are drawn so that they intersect one of the circles at the points A and B, and the other circle at the points C and D respectively, let us prove that ∠AQC = ∠BQD. ​

Answer 1

Answer:

Solution :-

Given :-

• Two circles with centre K and L intersect at the points P and Q. Through the point P a straight line is drawn which intersects one circle at the points A and B and other circles at the points C and D.

We have to prove ∠AQC = ∠BQD.

Construction :-

• A, Q; B, Q; C,Q and D, Q are joined.

Proof :-

∠PCQ and ∠PDQ are the angles the same arc,

SO,

∠PCQ = ∠PDQ . . . . . . (1)

Again, ∠PAQ and ∠PBQ are the same circular angle,

So,

∠PAQ = ∠PBQ . . . . . . . . (2)

Adding (1) and (2)

∴ ∠PAQ + ∠PCQ = ∠PBQ + ∠P . . . . . (3)

Sum of three angles are = 180°

∴ ∠CAQ + ∠ACQ + ∠AQC = 180°

or, ∠PAQ + ∠PCQ + ∠ACQ = 180° . . . . (4)

Similarly, ∠DBQ + ∠BDQ + ∠BCQ = 180°

or, ∠PBQ + ∠PDQ + ∠BQD = 180° . . . . (5)

From (4) and (5)

∠PAQ + ∠DCQ + ∠AQC = ∠PBQ + ∠PDQ + ∠BQD . . . . (6)

Substracting (3) from (4) ∠ACQ = ∠BQD (Proved).