Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.
To find : ∠APB
Proof: let ∠CAP = α and ∠CBP = β.
CA = CP [lengths of the tangents from an external point C]
In a triangle PAC, ∠CAP = ∠APC = α
similarly CB = CP and ∠CPB = ∠PBC = β
now in the triangle APB,
∠PAB + ∠PBA + ∠APB = 180° [sum of the interior angles in a triangle]
α + β + (α + β) = 180°
2α + 2β = 180°
α + β = 90°
∴ ∠APB = α + β = 90°