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)