A circle touches the side BC of a triangle ABC at P and the extended sides AB and AC at Q and R respectively. Prove that AQ = (BC+CA+AB)

Asked by Topperlearning User | 8th Dec, 2013, 12:37: AM

Expert Answer:

BQ = BP (lengths of tangents drawn from an external point to a circle are equal)

Similarly, CP=CR , and AQ=AR

2AQ = AQ+AR

= (AB+BQ) + (AC+CR)

= AB+BP+AC+CP

= (BP+CP) +AC+AB

2AQ = BC+CA+AB

AQ = (BC+CA+AB)

Answered by  | 8th Dec, 2013, 02:37: AM