# Two equal circles

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2: Sum of all angles of a straight line = 180°.

Property 3: Sum of all angles of a triangle = 180°.

By property 1, ∆OAB is right-angled at OAB (i.e., OAB = 90°) and PBA is right-angled at PBA (i.e., PBA = 90°)

Clearly,

b + c = OAB

b + c = 90°

b = 90° - c

Similarly,

d + e = PBA

d + e = 90°

e = 90° - d

Now,

a = b = 90° - c [ OA = OC (Radius)]

And,

e = f = 90° - d [ PB = PC (Radius)]

By property 2,

a + f + ACB = 180°

ACB = 180° a f

ACB = 180° (90° - c) (90° - d)

ACB = 180° 90° + c 90° + d

ACB = c + d

Now, in ∆ACB

By property 3,

ACB + c + d = 180°

ACB + ACB = 180° [∵∠ACB = c + d]

2ACB = 180°

ACB = 90°

Hence, ACB = 90°

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