If-in-ABC-r-1-r-2-r-3-r-prove-that-triangle-is-right-angled-

Question Number 16273 by Tinkutara last updated on 20/Jun/17

If in Δ A B C r_{1} = r_{2} + r_{3} + r, prove

that triangle is right angled .

Commented by mrW1 last updated on 20/Jun/17

what is r, r_{1}, \dots ?

Commented by Tinkutara last updated on 20/Jun/17

r is the inradius of Δ A B C .

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

opposite to angles A, B and C

respectively .

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

p . t g \frac{A}{2} = p . t g \frac{B}{2} + p . t g \frac{C}{2} + \frac{S}{p}

\Rightarrow t g \frac{A}{2} = t g \frac{B}{2} + t g \frac{C}{2} + \frac{S}{p^{2}}

t g \frac{A}{2} . t g \frac{B}{2} . t g \frac{C}{2} = \frac{S}{p^{2}}

\Rightarrow t g \frac{A}{2} = t g \frac{B}{2} + t g \frac{C}{2} + t g \frac{A}{2} . t g \frac{B}{2} . t g \frac{C}{2}

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

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

\frac{A}{2} = \frac{B + C}{2} \Rightarrow A = B + C = 180 - A \Rightarrow A = 90^{∙} .

Commented by Tinkutara last updated on 20/Jun/17

Thanks Sir!

Answered by ajfour last updated on 20/Jun/17

r_{1} = \frac{Δ}{s - a}

r_{2} + r_{3} + r = \frac{Δ}{s - b} + \frac{Δ}{s - c} + \frac{Δ}{s}

w h e n a b o v e r_{1} = r_{2} + r_{3} + r

\frac{1}{s - a} = \frac{1}{s - b} + \frac{1}{s - c} + \frac{1}{s}

\frac{1}{s - a} - \frac{1}{s} = \frac{1}{s - b} + \frac{1}{s - c}

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

\frac{a}{s (s - a)} = \frac{a}{(s - b) (s - c)}

o r (s - b) (s - c) = s (s - a)

b c - s (b + c) = - a s

b c = s (b + c - a)

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

2 b c = (b + c + a) (b + c - a)

2 b c = b^{2} + b c - a b + b c + c^{2} - a c + a b

+ a c - a^{2}

\Rightarrow a^{2} = b^{2} + c^{2} .

Commented by Tinkutara last updated on 20/Jun/17

Thanks Sir!

Answered by mrW1 last updated on 20/Jun/17

with r_{1} = r_{2} + r_{3} + r we have

\frac{Δ}{s - a} = \frac{Δ}{s - b} + \frac{Δ}{s - c} + \frac{Δ}{s}

\frac{1}{s - a} - \frac{1}{s} = \frac{1}{s - b} + \frac{1}{s - c}

\frac{a}{s (s - a)} = \frac{2 s - b - c}{(s - b) (s - c)}

\frac{a}{s (s - a)} = \frac{a}{(s - b) (s - c)}

\Rightarrow s (s - a) = (s - b) (s - c)

sa - s (b + c) + bc = 0

s (a - b - c) + bc = 0

(a + b + c) (a - b - c) + 2 bc = 0

a^{2} - {(b + c)}^{2} + 2 bc = 0

a^{2} - b - c^{2} - 2 bc + 2 bc = 0

\Rightarrow a^{2} = b + c^{2}

\Rightarrow triangle is right angled!

Commented by Tinkutara last updated on 20/Jun/17

Thanks Sir!