prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the center
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In ΔOPA and ΔOCA
OP = OC .....Radii
AP = AC ......Tangents from point A
AO = AO (Common )
ΔOPA ≅ ΔOCA (By SSS criterion)
Therefore, ∠POA = ∠COA .... (1)
Similarly , ΔOQB ≅ ΔOCB
∠QOB = ∠COB .......(2)
POQ is a diameter of the circle.
Hence, it is a straight line.
Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°
From equations (1) and (2), it can be observed that
2∠COA + 2∠COB = 180°
∴ ∠COA + ∠COB = 90°
∴ ∠AOB = 90°.
OP = OC .....Radii
AP = AC ......Tangents from point A
AO = AO (Common )
ΔOPA ≅ ΔOCA (By SSS criterion)
Therefore, ∠POA = ∠COA .... (1)
Similarly , ΔOQB ≅ ΔOCB
∠QOB = ∠COB .......(2)
POQ is a diameter of the circle.
Hence, it is a straight line.
Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°
From equations (1) and (2), it can be observed that
2∠COA + 2∠COB = 180°
∴ ∠COA + ∠COB = 90°
∴ ∠AOB = 90°.
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hope helped
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