In-ABC-r-1-r-2-and-r-3-are-the-exradii-as-shown-Prove-that-r-1-s-a-r-2-s-b-and-r-3-s-c-Here-s-a-b-c-2-

Question Number 16214 by Tinkutara last updated on 19/Jun/17

In Δ A B C, r_{1}, r_{2} and r_{3} are the exradii

as shown . Prove that r_{1} = \frac{Δ}{s - a},

r_{2} = \frac{Δ}{s - b} and r_{3} = \frac{Δ}{s - c} . Here

s = \frac{a + b + c}{2} .

Commented by Tinkutara last updated on 19/Jun/17

Commented by Tinkutara last updated on 20/Jun/17

Thanks Sir!

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

S = a r e a o f Δ A B C, p = s e m i p r e m e t e e .

C E = r_{1} . c o t g (\frac{π}{2} - \frac{C}{2}) = r_{1} . t g \frac{C}{2}

t g \frac{A}{2} = \frac{r_{1}}{A E} \Rightarrow r_{1} = A E . t g \frac{A}{2} = (b + r_{1} . t g \frac{C}{2}) . t g \frac{A}{2}

\Rightarrow r_{1} (1 - t g \frac{A}{2} . t g \frac{C}{2}) = b . t g \frac{A}{2}

\Rightarrow r_{1} = \frac{b . t g \frac{A}{2}}{1 - t g \frac{A}{2} . t g \frac{C}{2}}

t g \frac{A}{2} = \frac{s i n \frac{A}{2}}{c o s \frac{A}{2}} = \frac{\sqrt{\frac{1 - c o s A}{2}}}{\sqrt{\frac{1 + c o s A}{2}}} = \sqrt{\frac{1 - \frac{b^{2} + c^{2} - a^{2}}{2 b c}}{1 + \frac{b^{2} + c^{2} - a^{2}}{2 b c}}} =

= \sqrt{\frac{a^{2} - {(b - c)}^{2}}{{(b + c)}^{2} - a^{2}}} = \sqrt{\frac{(a + c - b) (a + b - c)}{(b + c + a) (b + c - a)}} =

= \sqrt{\frac{2 (p - b) . 2 (p - c)}{2 p . 2 (p - a)}} = \sqrt{\frac{(p - b) (p - c)}{p (p - a)}}

\Rightarrow r_{1} = \frac{b . \sqrt{\frac{(p - b) (p - c)}{p (p - a)}}}{1 - \sqrt{\frac{(p - b) (p - c)}{p (p - a)} . \frac{(p - a) (p - b)}{p (p - c)}}} =

= \frac{b . \frac{\sqrt{p (p - a) (p - b) (p - c)}}{p (p - a)}}{1 - \frac{p - b}{p}} = \frac{\frac{b . S}{p (p - a)}}{\frac{p - (p - b)}{p}} =

= \frac{b . S}{b (p - a)} = \frac{S}{p - a} .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Jun/17

A D = A E = b + r_{1} . t g \frac{C}{2}

{D E}^{2} = r_{1}^{2} + r_{1}^{2} - 2 r_{1}^{2} c o s (180 - A) =

= 2 r_{1}^{2} (1 + c o s A) = 4 r_{1}^{2} {c o s}^{2} \frac{A}{2} \Rightarrow

D E = 2 r_{1} . c o s \frac{A}{2}

{D E}^{2} = {A D}^{2} + {A D}^{2} - 2 {A D}^{2} . c o s A =

= 2 {A D}^{2} (1 - c o s A) = 4 {A D}^{2} . {s i n}^{2} \frac{A}{2}

\Rightarrow D E = 2 A D . s i n \frac{A}{2} = 2 r_{1} . c o s \frac{A}{2} \Rightarrow

b + r_{1} . t g \frac{C}{2} = r_{1} . c o t g \frac{A}{2} \Rightarrow r_{1} = \frac{b . t g \frac{A}{2}}{1 - t g \frac{A}{2} . t g \frac{C}{2}}

t g \frac{A}{2} = \frac{S}{p (p - a)}, t g \frac{A}{2} . t g \frac{C}{2} = \frac{p - b}{p}

\Rightarrow r_{1} = \frac{\frac{b . S}{p (p - a)}}{1 - \frac{p - b}{p}} = \frac{b . S}{b . (p - a)} = \frac{S}{p - a} .

n o t e : r_{1} = p . t g \frac{A}{2}, r_{2} = p . t g \frac{B}{2}, r_{3} = p . t g \frac{C}{2}

t g \frac{A}{2} . t g \frac{B}{2} . t g \frac{C}{2} = \sqrt{\frac{(p - b) (p - c) (p - a) (p - c) (p - a) (p - b)}{p (p - a) . p (p - b) . p (p - c)}} = \frac{S}{p^{2}}

\Rightarrow S = p^{2} . t g \frac{A}{2} . t g \frac{B}{2} . t g \frac{C}{2} = p^{2} . \frac{r_{1}}{p} . \frac{r_{2}}{p} . \frac{r_{3}}{p}

\Rightarrow S . p = r_{1} . r_{2} . r_{3} \Rightarrow S . (p . r) = r . r_{1} . r_{2} . r_{3}

\Rightarrow S^{2} = r . r_{1} . r_{2} . r_{3} \Rightarrow S = \sqrt{r . r_{1} . r_{2} . r_{3}} .

Answered by ajfour last updated on 19/Jun/17

Commented by Tinkutara last updated on 22/Jun/17

Then this procedure is wrong ? Or how

to proceed further using this method ?

Commented by ajfour last updated on 22/Jun/17

l e n g t h s o f t a n g e n t f r o m B t o

e x c i r c l e o f r a d i u s r_{1} a r e

B D = B E = m a

l e n g t h s o f t a n g e n t f r o m C t o

e x c i r c l e o f r a d i u s r_{1} a r e

C D = C F = n a

m a + n a = a

\Rightarrow m + n = 1 \dots \dots (i)

I f Δ i s t h e a r e a o f △ A B C,

A r e a o f e n t i r e f i g u r e (c o n s i d e r

A G j o i n e d (D m a y o r m a y n o t

l i e o n i t) i s

A = Δ + 2 [\frac{r_{1}}{2} (m a) + \frac{r_{1}}{2} (n a)]

= Δ + r_{1} (m a + n a)

A = Δ + r_{1} a (a s m a + n a = a)

\dots . (i i)

b u t A i s a l s o g i v e n b y

A = \frac{1}{2} (r_{1}) (c + m a) + \frac{1}{2} (r_{1}) (b + n a)

= \frac{r_{1}}{2} (b + c + m a + n a)

A = \frac{r_{1}}{2} (a + b + c) (a s m + n = 1)

\dots . (i i i)

e q u a t i n g (i i) a n d (i i i) :

A = \frac{r_{1}}{2} (a + b + c) = Δ + r_{1} a

o r r_{1} s - r_{1} a = Δ

r_{1} = \frac{Δ}{s - a} w h e r e s = \frac{a + b + c}{2} .

s i m i l a r l y r_{2} = \frac{Δ}{s - b} a n d r_{3} = \frac{Δ}{s - c}

i a m t h i n k i n g h o w t o p r o v e . .

Commented by mrW1 last updated on 19/Jun/17

Excellent sir!

Commented by ajfour last updated on 22/Jun/17

e r r o r w r i t i n g e q n . (i) b u t i h a v e

u s e d m a + n a = a, i t s e l f, y o u c a n

s e e f o r y o u r s e l f .

Answered by mrW1 last updated on 20/Jun/17

Proof in an other way :

The incircle of Δ ABC has the radius

r_{i} = \frac{Δ}{s}

with Δ = area of Δ ABC and

s = \frac{a + b + c}{2}

When we extend B and C to B^{'} and C^{'},

with B^{'} C^{'} ∣∣ BC, see diagram, then

the excircle r_{1} becomes the incircle

of the new triangle Δ {AB}^{'} C^{'} .

Commented by mrW1 last updated on 19/Jun/17

Commented by mrW1 last updated on 20/Jun/17

Height of Δ ABC is h_{A} = AD

since Δ = \frac{1}{2} {ah}_{A}

\Rightarrow h_{A} = \frac{2 Δ}{a}

Height of Δ {AB}^{'} C^{'} is h_{A^{'}} = {AD}^{'}

h_{A^{'}} = h_{A} + {2 r}_{1}

since Δ {AB}^{'} C \sim Δ ABC, we have

\frac{r_{1}}{r_{i}} = \frac{h_{A^{'}}}{h_{A}} = \frac{h_{A} + {2 r}_{1}}{h_{A}} = 1 + \frac{{2 r}_{1}}{h_{A}} = 1 + \frac{{ar}_{1}}{Δ}

r_{1} Δ = r_{i} Δ + {ar}_{1} r_{i}

(\frac{Δ}{r_{i}} - a) r_{1} = Δ

(s - a) r_{1} = Δ

\Rightarrow r_{1} = \frac{Δ}{s - a}

similarly

\Rightarrow r_{2} = \frac{Δ}{s - b}

\Rightarrow r_{3} = \frac{Δ}{s - c}

additionally we get :

{rr}_{1} r_{2} r_{3} = \frac{Δ^{4}}{s (s - a) (s - b) (s - c)} = \frac{Δ^{4}}{Δ^{2}} = Δ^{2}

\Rightarrow Δ = \sqrt{{rr}_{1} r_{2} r_{3}}

Commented by Tinkutara last updated on 20/Jun/17

Thanks Sir!