# Properties of Triangles – EX-RADII

## Properties of Triangles – EX-RADII

Definition: The internal bisector of angle A and external bisectors of angles B, C of ΔABC are concurrent. The point of concurrence is called ex-centre of ΔABC opposite to the vertex A. It is denoted by I₁. The point I₁ is equidistant to the sides of the triangle. The circle with centre I₁ and touching the three sides of the triangle is called excircle of triangle ABC opposite to the vertex A. The radius of this ex-circle is called ex-radius of triangle ABC and it is denoted by r₁. The excentres of ΔABC opposite to the vertices B, C are respectively denoted by I₂, I₃. The ex-radii of triangle ABC opposite to the vertices B, C are respectively denoted by r₂, r₃.

Statement: In ΔABC, $${{r}_{1}}=\frac{\Delta }{s-a}$$.

Proof: Let I₁ be the excentre of ΔABC opposite to the vertex A.

Let X, Y, Z be the projections of I₁ on BC, CA, AB. ∴ I₁X = I₁Y = I₁Z = r₁

Δ = Area of ΔABC

= Area of Δ AI₁B + Area of ΔAI₁C – Area of ΔBI₁C

= ½ AB.I₁X + ½AC.I₁Y – ½BC.I₁Z = ½c r₁ + ½b r₁ – ½a r₁

= ½r₁ (c + b – a) = ½r₁ 2(s – a) = r₁ (s – a) → r₁ = Δ / s – a.

Similarly, r₂ = Δ / s – b, r₃ = Δ / s – c.

Statement: In ΔABC, r₁ = 4R sinA/2 cos B/2 cos C/2.

Proof: 4R sin A/2 cos B/2 cos C/2$$=4R\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{s(s-b)}{ca}}\sqrt{\frac{s(s-c)}{ab}}$$,

$$=\frac{4Rs\left( s-b \right)\left( s-c \right)}{abc}=\frac{4Rs\left( s-a \right)\left( s-b \right)\left( s-c \right)}{\left( s-a \right)4R\Delta }=\frac{{{\Delta }^{2}}}{\left( s-a \right)\Delta }=\frac{\Delta }{s-a}={{r}_{1}}$$.

Statement: In $$\Delta ABC,\,{{r}_{1}}=s\tan \frac{A}{2}=\left( s-b \right)\cot \frac{C}{2}=\left( s-c \right)\cot \frac{B}{2}$$.

Proof: $$\Delta ABC,\,{{r}_{1}}=s\tan \frac{A}{2}=\left( s-b \right)\cot \frac{C}{2}=\left( s-c \right)\cot \frac{B}{2}$$

$$\left( s-b \right)\cot \frac{C}{2}=\left( s-b \right)\sqrt{\frac{s\left( s-c \right)}{\left( s-a \right)\left( s-b \right)}}=\frac{\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}}{s-a}=\frac{\Delta }{s-a}={{r}_{1}}$$

Similarly, $$\left( s-c \right)\cot \frac{B}{2}={{r}_{1}}$$.

Statement: In$$\Delta ABC,\,{{r}_{1}}=\frac{a}{\tan \left( B/2 \right)+\tan \left( C/2 \right)}$$.

Proof: $${{r}_{1}}\left( \tan \frac{B}{2}+\tan \frac{C}{2} \right)=\frac{\Delta }{s-a}\left[ \sqrt{\frac{\left( s-c \right)\left( s-a \right)}{s\left( s-b \right)}}+\sqrt{\frac{\left( s-a \right)\left( s-b \right)}{s\left( s-c \right)}} \right]$$

$$=\frac{\Delta }{s-a}\left[ \frac{\left( s-c \right)\sqrt{s-a}+\left( s-b \right)\sqrt{s-a}}{\sqrt{s\left( s-b \right)\left( s-c \right)}} \right]=\frac{\Delta \sqrt{s-a}\left( s-c+s-b \right)}{\left( s-a \right)\sqrt{s\left( s-b \right)\left( s-c \right)}}$$

$$=\frac{\Delta \left( a+b+c-c-b \right)}{\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}}=\frac{\Delta a}{\Delta }=a$$

∴ $${{r}_{1}}=\frac{a}{\tan \left( B/2 \right)+\tan \left( C/2 \right)}$$.

Statement: In ΔABC,

i) $$\frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}}=\frac{1}{r}$$

ii) rr₁r₂r₃ = Δ²

Proof:

i) $$\frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}}=\frac{s-a}{\Delta }+\frac{s-b}{\Delta }+\frac{s-c}{\Delta }=\frac{s-a+s-b+s-c}{\Delta }$$

$$=\frac{3s-\left( a+b+c \right)}{\Delta }=\frac{3s-2s}{\Delta }=\frac{s}{\Delta }=\frac{1}{r}$$.

ii) $$r{{r}_{1}}{{r}_{2}}{{r}_{3}}=\frac{\Delta }{s}.\frac{\Delta }{s-a}\frac{\Delta }{s-b}\,.\,\frac{\Delta }{s-c}=\frac{{{\Delta }^{4}}}{{{\Delta }^{2}}}={{\Delta }^{2}}$$.