In the given figure, the incircle of △ABC touches the sides BC,CA, and AB at P,Q, and R respectively.
Prove that (AR+BP+CQ)=(AQ+BR+CP)=12(Perimeter of △ABC).
Answer:
- 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). - 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)
- Thus, we can say that (AR+BP+CQ)=(AQ+BR+CP)=12 (Perimeter of △ABC).