# A circle touches the side BC of ΔABC at P, AB and AC produced at Q and R respectively, prove that AQ = AR = ( perimeter of ΔABC) Lengths of tangents drawn from an exterior point to the circle are equal.

From the above theorem,

AR = AQ …(1)

BP = BQ …(2)

CP = CR …(3)

Now,

Perimeter of ΔABC = AB + BC + CA

Perimeter of ΔABC = AB + BP + PC + CA

Perimeter of ΔABC = (AB + BQ) + (CR + CA) [Using (2) and (3)]

Perimeter of ΔABC = AQ + AR

Perimeter of ΔABC = AQ + AQ

Perimeter of ΔABC = 2AQ [Using (1)] AQ = AR = 1/2Perimeter of ΔABC

Hence, Proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Section Formula30 mins  Previous Year RMO Questions43 mins  NCERT | Most Important Proofs for Boards28 mins  Know About Important Proofs in Triangles33 mins  Champ Quiz | Previous Year NTSE QuestionsFREE Class  Measuring distance by Distance formula49 mins  Champ Quiz | Distance Formula30 mins  Imp. Qs. on Distance Formula68 mins  Coordinate Geometry Important Questions38 mins  Geometry of Temples36 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 