If a circle touches the side \(BC\) of a triangle \(ABC\) at \(P\) and extended sides \(AB\) and \(AC\) at \(Q\) and \(R\), respectively. Prove that \(AQ = \frac{1}{2} (BC+CA+AB)\).
Lengths of two tangents drawn from an external point to a circle are equal.
\(BQ = \) - - - - (i)
\(PC = \) - - - - (ii)
\(AQ = \) - - - - (iii)
From (iii), we have
\(AB + BQ = AC + CR\)
\(AB + BP = AC + \) (using (i) and (ii))
of \(\Delta ABC = AB + BC + AC\)
\(= AB + (BP + PC) + AC\)
Simplyfing this then we get,
\(AB+BC+AC= 2 (AB + )\)
\(= 2 (AB + BQ)\)
\(= 2 AQ\)
From this we get, \(AQ = \frac{1}{2} (AB + BC + CA)\)
Hence proved.