Question

two circles touch each other externally at a point c and p is a point on the common tangent at c if pA and PB are tangents to two circles prove that pa equals to pb

Answer 1

Answer:

sorry I didn't understand your question

Answer 2

Answer:

Step-by-step explanation:

Don't focus on my drawing ..it's poor lol

O1 and O2 are centers of circles.

In triangle BPO1 and CPO1

angle PBO1 =90  (tangent make 90 with radius of circle)

angle PCO1=90  (  same reason)

PO1 =PO1 (hypotenuse of both triangle)

BO1 =CO1 (radius )

So, triangle BPO1 and CPO1 are congruent by RHS (right angle,hypotenuse,side) property

Hence, BP = CP

Similarly, trianlge PCO2 and PAO2 are also congruent

and HEnce , AP = CP

since, BP =CP and AP =CP

we can say BP =AP

Proved..

Hope it helps :-)