 Hello and welcome to the session, let us discuss the following question. It says, a circle touches the side BC of triangle ABC at P and touch AB and AC when produced at Q and R respectively as shown in figure. Show that AQ is equal to half the perimeter of triangle ABC. So let us now move on to the solution and let's first write what is given to us. We are given that a circle touches the side BC of triangle ABC when produced at Q and R respectively. AQ is equal to half the perimeter of triangle ABC. Let's now start the proof. AQ is equal to AR also BP is equal to BQ and similarly CP is equal to CR. This is because tangents drawn on a point to a circle are equal. This name this as 1, this as 2 and this as 3. Now the perimeter of the triangle ABC given by some of the lengths of the sides of the triangle that is AB plus BC then equal to AB plus BC. Now BC can be written as BP plus BC. Now again this is equal to AB plus. Now BP is same as BQ same as CR E plus BQ is equal to AQ E plus CR is AR. That is this. From 1 we know that AQ is equal to AR. This is AQ plus AQ and this is by using and this is from this is by 2 and 3 and this is from 1. AQ plus AQ is 2AQ. So we have 2AQ is equal to the perimeter of triangle ABC. So this implies AQ is equal to 1 by 2 of the perimeter by angle ABC. And this is what we had to prove. So hence the result is proved. So this completes the question and the session. Bye for now. Take care and have a good day.