A circle is touching the side BC of a Δ ABC at P and is touching AB and AC when produced at Q and R respectively.

Question

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

Philoid · Accepted Answer

In the given figure,

AQ and AR are two tangents drawn from an exterior point A at contact points Q and R on the circle.

∴ AQ = AR

⇒ AQ = AC + CR…..(1)

Similarly,

BQ and BP are two tangents drawn from an exterior point B at contact points Q and P on the circle.

∴ BQ = BP…..(2)

And,

CR and CP are two tangents drawn from an exterior point C at contact points R and P on the circle.

∴ CR = CP…..(3)

Now, Equation (1) can be written as –

AQ = (AC + CR + AC + CR)/2

⇒ AQ = (AC + CP + AC + CR)/2 [using(3)]

⇒ AQ = (AC + CP + AR)/2

⇒ AQ = (AC + CP + AQ)/2

⇒ AQ = (AC + CP + AB + BQ)/2

⇒ AQ = (AC + CP + AB + BP)/2 [using(2)]

⇒ AQ = (AB + BC + AC)/2 [∵BP + CP=BC]

Thus, AQ = (1/2) × perimeter of Δ ABC

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

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