Question

this is my question

Answer 1

Perimeter of ∆ = AB+CA +BC= (AQ-BQ ) +(AR-CR) +( BP+CP)= AQ+AR. { Since BP=BQ AND CP=CR as per the theorem " the length of the tangents drawn from a point are equal"}

Again AQ=AR( for the same reason)
So AQ+AR= 2AQ= perimeter of∆ABC

Answer 2

It was quite simple
Hope you may understand the solution☺☺☺☺☺☺