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

To prove that AQ = Perimeter of triangle ABC, we need to understand a few properties of circles and triangles.

Step 1: Understand the problem In triangle ABC, a circle is inscribed such that it touches BC at point P, AB extended at point Q, and AC extended at point R. We need to prove that the length of AQ is equal to the perimeter of triangle ABC.

Step 2: Use the tangent properties From the properties of a circle, we know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore, we have AP = AQ and CP = CR.

Step 3: Use the triangle properties In triangle ABC, the side BC can be expressed as the sum of BP and PC. Similarly, AB = AP + BP and AC = AP + PC.

Step 4: Substitute the values Substituting the values from step 2 in step 3, we get AB = AQ + BP and AC = AQ + PC.

Step 5: Find the perimeter The perimeter of triangle ABC is given by AB + BC + AC. Substituting the values from step 4, we get Perimeter = AQ + BP + AQ + PC + AQ + BP + PC.

Step 6: Simplify the equation Simplifying the equation from step 5, we get Perimeter = 3AQ.

Step 7: Prove the equality Since AQ is not equal to zero, we can divide both sides of the equation by 3 to get AQ = Perimeter of triangle ABC.

Therefore, we have proved that AQ is equal to the perimeter of triangle ABC.