Properties of Triangles – ExCircle
If r is radius of in circle and r₁, r₂, r₃ are the radii of ex-circle opposite to the vertices A, B, C of ΔABC respectively then
(i) \(r=\frac{\Delta }{s}\).
\({{r}_{1}}=\frac{\Delta }{s-a}\).
\({{r}_{2}}=\frac{\Delta }{s-b}\).
\({{r}_{3}}=\frac{\Delta }{s-c}\).
(ii) \(r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\).
\({{r}_{1}}=4R\sin \frac{A}{2}.\cos \frac{B}{2}.\cos \frac{C}{2}\).
\({{r}_{2}}=4R\cos \frac{A}{2}.\sin \frac{B}{2}.\cos \frac{C}{2}\).
\({{r}_{3}}=4R\cos \frac{A}{2}.\cos \frac{B}{2}.\sin \frac{C}{2}\).
(iii) \(r=\left( s-a \right).\tan \frac{A}{2}=(s-b).\tan \frac{B}{2}=(s-c).\tan \frac{C}{2}\).
\({{r}_{1}}=s\tan \frac{A}{2}=(s-b)\cot \frac{C}{2}=(s-c)\cot \frac{B}{2}\).
\({{r}_{2}}=s\tan \frac{B}{2}=(s-c)\cot \frac{A}{2}=(s-a)\cot \frac{C}{2}\).
\({{r}_{3}}=s\tan \frac{C}{2}=(s-a)\cot \frac{B}{2}=(s-b)\cot \frac{A}{2}\).
Example: Show that ∑a cos A = 2 (R + r).
Solution: Given that
∑a cos A = 2 (R + r)
\(\sum{2R\ \sin A\ \frac{\cos A}{\sin A}}\),
∑2R cos A
\(2R(1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})\),
(since from transformation)
\(2(R+4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})\),
= 2(R + r).
Hence proved ∑a cos A = 2 (R + r).