Given X and Y are two circles touch each other externally at C. AB is the common tangent to the circles X and Y at point A and B respectively.

To find: ∠ACB

Proof: Let P be a point on AB such that, PC is at right angles to the line joining the centers of the circles.

Note that, PC is a common tangent to both circles.

This is because the tangent is perpendicular to the radius at the point of contact for any circle.

Let ∠PAC = α and ∠PBC = β.

PA = PC [lengths of the tangents from an external point C]

In a triangle CAP, ∠PAC = ∠ACP = α

Similarly, PB = CP and ∠PCB = ∠CBP = β.

Now in the triangle ACB:

∠CAB + ∠CBA + ∠ACB = 180° [sum of the interior angles in a triangle]

α + β + (α + β) = 180° (Since ∠ACB = ∠ACP + ∠PCB = α + β).

2α + 2β = 180°

⇒  α + β = 90°

∴ ∠ACB = α + β = 90°