# Properties of Triangles – ExCircle

## 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).