A circle touches the side BC of a triangle ABC at P and touches AB and AC when produced at Q and R.

Show that AQ= 1/2 (perimeter of triangle ABC)

Ans:    Since the length of tangents from external point to a circle are equal.

AQ = AR BQ = BP PC = CR

Since  AQ = AR

AB + BQ = AC + CR

∴ AB + BP = AC + PC (Since BQ = BP & PC = CR)

Perimeter of ? ABC = AB + AC + BC

= AB + BP + PC + AC

= AQ + PC + AC (Since AB + BP = AQ)

= AQ + AB + BP (Since PC + AC = AB + BP)

= AQ + AQ (Since AB + BP = AQ)

Perimeter of ? ABC = 2AQ

∴ AQ = 1/2 (perimeter of triangle ABC)