In figure OP is equal to diameter of the circle. AP and PB are two tangents to the circle. AB is a secant to the circle. Prove that triangle ABP is a equilateral triangle.
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Join A to B.
We have
OP= Diameter
Or, OQ + OP= diameter
Or, Radius + PQ = diameter (that is OQ=radius)
Or, PQ = diameter – radius
Or, PQ = radius
Or, OQ = PQ = radius
Thus OP is the hypotenuse of the right angle triangle AOP.
So, in ∆AOP sin of angle P = OP /AO
= 1/2
So, P = 30°
Hence, APB = 60°
Now, As in ∆AOP, AP=AB
So, PAB = PBA =60°
Hence,
∆ABP is equilateral triangle.
We have
OP= Diameter
Or, OQ + OP= diameter
Or, Radius + PQ = diameter (that is OQ=radius)
Or, PQ = diameter – radius
Or, PQ = radius
Or, OQ = PQ = radius
Thus OP is the hypotenuse of the right angle triangle AOP.
So, in ∆AOP sin of angle P = OP /AO
= 1/2
So, P = 30°
Hence, APB = 60°
Now, As in ∆AOP, AP=AB
So, PAB = PBA =60°
Hence,
∆ABP is equilateral triangle.
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