Two congruent circles intersect each other at O. Through A any line segment PAQ is drawn so that P and Q lie on the circle. Prove that BP = BQ.
Maybe the circles intersect at A and B
Ya then it's very easy
The same sum is given in my textbook and it's written that circles intersect at A and B and P and Q lie on the two circles.
I will write the answer accordingly
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Answered by
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See the pic for reference.
Chord AB will subtend equal angles on both the circles' circumferences.
So, Angle APB = Angle AQB
So, in triangle BPQ,
QB = PB (Because they are sides opposite to equal angles)
Hence proven.
Chord AB will subtend equal angles on both the circles' circumferences.
So, Angle APB = Angle AQB
So, in triangle BPQ,
QB = PB (Because they are sides opposite to equal angles)
Hence proven.
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Answered by
40
AB is a common cord of congruent circles
.'. angle subtended by that chord are equal
.'. angle AQB = angle APB
now in ΔPQB
'.' angle AQB = angle APB
.'. BQ = BP ( because sides opposite to equal angles are equal)
.'. angle subtended by that chord are equal
.'. angle AQB = angle APB
now in ΔPQB
'.' angle AQB = angle APB
.'. BQ = BP ( because sides opposite to equal angles are equal)
Attachments:
please mark it as best answer
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But where are points A and B? P and Q lie on which circle?