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Question

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.


Solution

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]

x+y+(x+y)=180

2x+2y=180

x+y=90

therefore ∠APB=x+y=90

hence AB subtends a right angle at P


Mathematics
Concise Mathematics
Standard X

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