Question
Wed March 06, 2013 By:
a circle is circumscribed in a triangle ABC. the sides AB ,BC ,CA of the triangle touches the circle at points P, Q,R respectively
prove that :
i)AB + CQ =AC + BQ

see below

Expert Reply
Wed March 06, 2013

As PB and BQ are tangents from B to the circle   so

PB=BQ

Similarly

AP=AR

CR=CQ

AP+PB+CQ=AR+BQ+CR

AB+CQ=AC+BQ

Also

Area of triangle=1/2(AB+BC+AC)*r

                            =1/2(2s)*r

                          = ½(perimeter)*r

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