A circle is touching the side BCBC of △ABC△ABC at PP and touching ABAB and ACAC produced at QQ and RR respectively. Prove that AQ=12(Perimeter of △ABC)AQ=12(Perimeter of △ABC).
Answer:
- We know that the lengths of tangents drawn from an external point to a circle are equal.
Thus, AQ=AR…(i)[Tangents from A]BP=BQ…(ii)[Tangents from B]CP=CR…(iii)[Tangents from C] - We know that the perimeter of a triangle is the sum of the lengths of its sides. So, perimeter of △ABC=AB+BC+AC=AB+(BP+CP)+AC[As, BC = BP + CP]=AB+BQ+CR+AC[Using eq(ii) and eq(iii)]=AQ+AR[As, AQ = AB + BQ and AR = CR + AC] =2AQ[Using eq(i)]
- Therefore, AQ=12(Perimeter of △ABC).