In the given figure, the incircle of ABCABC touches the sides BC,CA,BC,CA, and ABAB at P,Q,P,Q, and RR respectively.
Prove that (AR+BP+CQ)=(AQ+BR+CP)=12(Perimeter of ABC).(AR+BP+CQ)=(AQ+BR+CP)=12(Perimeter of ABC).
R Q P B C A


Answer:


Step by Step Explanation:
  1. We know that the lengths of tangents from an external point to a circle are equal.
    Thus, AR=AQ(i)[Tangents from A] BP=BR(ii)[Tangents from B] CQ=CP(iii)[Tangents from C]  Adding (i),(ii), and (iii) equations, we have AR+BP+CQ=AQ+BR+CP=k(say).
  2. We know that the perimeter of a triangle is the sum of its sides.  So, perimeter of ABC=AB+BC+CA=(AR+BR)+(BP+CP)+(CQ+AQ)=(AR+BP+CQ)+(AQ+BR+CP)=(k+k)=2k Thus, k=12 (Perimeter of ABC)
  3. Thus, we can say that (AR+BP+CQ)=(AQ+BR+CP)=12 (Perimeter of ABC).

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