In the given figure, the incircle of △ABC△ABC△ABC touches the sides BC,CA,BC,CA,BC,CA, and ABABAB at P,Q,P,Q,P,Q, and RRR respectively.
Prove that (AR+BP+CQ)=(AQ+BR+CP)=12(Perimeter of △ABC).(AR+BP+CQ)=(AQ+BR+CP)=12(Perimeter of △ABC).(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, [Math Processing Error] Adding (i),(ii),(i),(ii), and (iii)(iii) equations, we have AR+BP+CQ=AQ+BR+CP=k(say).AR+BP+CQ=AQ+BR+CP=k(say). - We know that the perimeter of a triangle is the sum of its sides. [Math Processing Error] Thus, k=12 (Perimeter of △ABC)k=12 (Perimeter of △ABC)
- Thus, we can say that (AR+BP+CQ)=(AQ+BR+CP)=12 (Perimeter of △ABC)(AR+BP+CQ)=(AQ+BR+CP)=12 (Perimeter of △ABC).