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Pythagoras Theorem

In a triangle...

Question

In a triangle $A B C,$ , prove that
$(\frac{b - c}{r_{1}}) + (\frac{c - a}{r_{2}}) + (\frac{a - b}{r_{3}}) = 0$

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Solution

We have $r_{1} = \frac{△}{s - a}$
$r_{2} = \frac{△}{s - b}$
and $r_{3} = \frac{△}{s - c}$
substituting the values of $r_{1}, r_{2}, r_{3}$ in the equation we get
$(\frac{b - c}{r_{1}}) + (\frac{c - a}{r_{2}}) + (\frac{a - b}{r_{3}})$
$= ⎛ ⎜ ⎜ ⎜ ⎝ \frac{b - c}{\frac{△}{s - a}} ⎞ ⎟ ⎟ ⎟ ⎠ + ⎛ ⎜ ⎜ ⎜ ⎝ \frac{c - a}{\frac{△}{s - b}} ⎞ ⎟ ⎟ ⎟ ⎠ + ⎛ ⎜ ⎜ ⎜ ⎝ \frac{a - b}{\frac{△}{s - c}} ⎞ ⎟ ⎟ ⎟ ⎠$
$= (\frac{b - c}{△}) (s - a) + (\frac{c - a}{△}) (s - b) + (\frac{a - b}{△}) (s - c)$
$= \frac{1}{△} [(b - c) (s - a) + (c - a) (s - b) + (a - b) (s - c)]$
$= \frac{1}{△} [s (b - c + c - a + a - b) - a (b - c) - b (c - a) - c (a - b)]$
$= 0$ (on simplification)

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Similar questions

Q. With usual notations, prove that in a triangle ABC

\frac{b - c}{r_{1}} + \frac{c - a}{r_{2}} + \frac{a - b}{r_{3}} = 0

Q. With usual notations, prove that in a triangle

A B C,

\frac{b - c}{r_{1}} + \frac{c - a}{r_{2}} + \frac{a - b}{r_{3}} = 0

Q. If in a triangle ABC , (a - b) (s - c) = (b - c) (s - a), prove that

r_{1}, r_{2}, r_{3}

Q. Prove that in a triangle

\frac{b c}{r_{1}} + \frac{c a}{r_{2}} + \frac{a b}{r_{3}} = 2 R [(\frac{a}{b} + \frac{b}{a}) + (\frac{b}{c} + \frac{c}{b}) + (\frac{c}{a} + \frac{a}{c}) - 3]

Q. Prove that :

a (r r_{1} + r_{2} r_{3}) = b (r r_{2} + r_{3} r_{1}) = c (r r_{3} + r_{1} r_{2})

r

is inradius and

r_{1}, r_{2}, r_{3}

are exradius of triangle

A B C

a, b, c

are the corresponding sides.

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