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, Adding and equations, we have - We know that the perimeter of a triangle is the sum of its sides. Thus,
- Thus, we can say that .