Q) A circle is touching the side BC of a Δ ABC at the point P and touching AB and AC produced at points Q and R respectively.
Prove that AQ = 1 (Perimeter of Δ ABC)

Ans:

Perimeter of the triangle Δ ABC, P = AB + AC + BC

∴ P = AB + AC + BP + CP              (since P ;lies on the line BC)

Now, BP and BQ are tangents on the circle from point B,

∴ BP = BQ

similarly, CP and CR are tangents from point C on the circle,

∴ CP = CR

Now, Perimeter of Δ ABC, P = AB + AC + BP + CP   

∴ P = AB + AC + (BQ) + (CR)

∴ P = (AB + BQ) + (AC +CR)

∴ P = AQ + AR

from the diagram, we can see that AQ and AR are tangents on the circle from point A,

∴ AQ = AR

Substituting this relation into equation of P, we get:

∴ P = AQ + AR = AQ + (AQ) = 2 A

∴ P = AQ + AR

∴ P = AQ + (AQ)

∴ P = 2 AQ

∴ AQ =

∴ AQ = (Perimeter of Δ ABC)

Hence Proved !

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