Question

The sides AB, BC and CA of triangle ABC touch a circle with center O and radius r at P, Q and R respectively.Prove that AB+CQ=AC+BQ and Area(ABC)=1/2(perimeter of ABC)

Answer 1

Step-by-step explanation:

By 10.2,

AP=AR

BP=BQ

CQ=CR

AB+CQ=(AP+PB)+CQ

=AP+BQ+CQ

=(AR+CQ)+BQ

=AC+BQ.

Area(triangle ABC)

=A(triangle OAB)+A(triangle OBC)+A(triangle OAC)

= 1/2×AB×r+1/2×BC×r+1/2×AC×r

=1/2×r(AB+BC+AC)

=1/2(PERIMETER OF TRIANGLE ABC)×r

Answer 2

Given:

AB, BC, and CA are sides of a triangle.

ABC touches center O

radius r at P, Q, and R respectively

To Prove:

AB + CQ = AC + BQ

Area(ABC) = 1/2 (Perimeter of ABC)

Solution:

The tangents to form an external point are equal.

So, AP = AR ..(i)

     BP = BQ ..(ii)

     CR = CQ ..(iii)

Then also,

          AB = AP + BP ..(iv)

          AC = AR + RC ..(v)

Now,

         AB = AP + BQ [ using (i), (ii), (iii), (iv), (v)]

         AC = AB - BQ ..(vi)

Also,

        AC = AP + CQ

        AP = AC - CQ ..(vii)

Using (vi) and (vii) ,

                 AB - BQ = AC - CQ

                 AB + CQ = AC + BQ

Hence, AC + CQ = AC + BQ

  For proving, area(ABC) = 1/2area(ABC), construct lines by joining OA, OB,       and OC.

OP = OQ = OR [ Radius of a circle].

Now,

     area(ABC) = area(ΔAOB) + area(ΔBOC) + area(AOC)

     area(ABC) = (1/2× AB ×OP) + (1/2× BC× OQ) + (1/2× AC× OR)

     area(ABC) = 1/2× r × ( AB + BC + AC)

     area(ABC) = 1/2 × r × (perimeter of ΔABC)

Hence, proved.