In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that :
(i) tangent at point P bisects AB,
(ii) angle APB = 90o.
given: C1 and C2 are two circles touch each other externally at P. AB is the common tangent to the circles C1 and C2 at point A and B respectively.
TPT:∠APB= 90 deg.
Proof: let ∠CAP=x and ∠CBP=y
CA=CP [lengths of the tangents from an external point C]
therefore in triangle PAC, ∠CAP=∠APC=x
similarly CB=CP and ∠CPB=∠PBC=y
now in the triangle APB,
∠PAB+∠PBA+∠APB=180 [sum of the interior angles in a triangle]
hence AB subtends a right angle at P