A circle is touching the side BC of triangle ABC at P and touching AB and AC produced at Q and R respectively. Prove that AQ = 1/2 (perimeter of triangle ABC)?





The first step to solve the problem is to draw a diagram. We have to draw a triangle ABC and a circle that touches the side BC at point P and touches the sides AB and AC produced at points Q and R respectively.



We know that the tangent to a circle is perpendicular to the radius at the point of contact. Therefore, we can draw perpendiculars from points Q and R to the line BC, as shown in the diagram. Let the length of the perpendicular from Q be x and the length of the perpendicular from R be y.



We know that the tangents from a point to a circle are equal in length. Therefore, we can write the following equations:

QA = QP + PA
RA = RP + PA

Since the circle is tangent to BC at P, we know that PA = PC. Therefore:

QA = QP + PC
RA = RP + PC



We know that the sum of the lengths of the sides of a triangle is equal to its perimeter. Therefore, we can write:

AB + BC + AC = perimeter of triangle ABC

We also know that BC = BP + PC. Therefore, we can write:

AB + BP + PC + AC = perimeter of triangle ABC



We know that the circle touches AB and AC produced at points Q and R respectively. Therefore, AQ = AB + x and AR = AC + y.



We can substitute the values of QA, RA, AQ, and AR into the equation we derived in step 4. This gives us:

AB + BP + QP + PC + AC + RP + PC = perimeter of triangle ABC

Simplifying this equation, we get:

AB + AC + 2PC + 2x + 2y = perimeter of triangle ABC

Substituting the values of AQ and AR from step 5, we get:

AB + AC + 2PC + AQ - AB + AR - AC = perimeter of triangle ABC

Simplifying this equation, we get:

2PC + 2x + 2y = perimeter of triangle ABC



We know that PC = (perimeter of triangle ABC)/2 - BC/2. Substituting this value into the equation we derived in step 6, we get:

(perimeter of triangle ABC)/2 - BC/2 + x + y = perimeter of triangle ABC/2

Simplifying this equation, we get:

x + y = (perimeter of triangle ABC)/2 - BC/2



We know that x and y are the lengths of the perpendiculars from Q and R to BC