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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

prove :

$\dfrac {1}{a+b}+\dfrac{1}{b+c}=\dfrac {3}{a+b+c}$

- Thread starter Albert
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- #1

- Jan 25, 2013

- 1,225

prove :

$\dfrac {1}{a+b}+\dfrac{1}{b+c}=\dfrac {3}{a+b+c}$

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- Feb 7, 2012

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Next, by the sine rule, $a,b,c$ are proportional to $\sin A,\sin B,\sin C$, so it will be sufficient to prove that $\sin^2B = \sin^2A + \sin^2C - \sin A\sin C.$

Also, we are told that $\angle A+\angle C =2 \angle B .$ But $\angle A+\angle C = 180^\circ - \angle B$, so that $3\angle B = 180^\circ$, $\angle B = 60^\circ$, and so $\sin^2 B = 3/4.$ And $\angle C = 120^\circ - \angle A$, so $\sin C = \sin120^\circ \cos A - \cos120^\circ \sin A = \frac{\sqrt3}2\cos A + \frac12\sin A.$ Therefore $$\begin{aligned}\sin^2A + \sin^2C - \sin A\sin C &= \sin^2A + \tfrac14(\sqrt3\cos A + \sin A)^2 - \tfrac12\sin A(\sqrt3\cos A + \sin A) \\ &= \sin^2A + \tfrac14(3\cos^2 A + 2\sqrt3\sin A\cos A + \sin^2 A) - \tfrac12\sin A(\sqrt3\cos A + \sin A) \\ &= \tfrac34\sin^2 A + \tfrac34\cos^2 A = \tfrac34 = \sin^2 B,\end{aligned}$$ as required.

- Jan 26, 2012

- 183

Couldn't you use use the law of Cosine with $\angle B = 60^\circ$ to get to the same answer quicker?

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- Feb 7, 2012

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Couldn't you use use the law of Cosine with $\angle B = 60^\circ$ to get to the same answer quicker?

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- Jan 25, 2013

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$b^2 = a^2 + c^2 - ac.$--------(1)

as jester said "Couldn't you use the law of Cosine with [FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]

using law of Cosine :

$b^2=a^2+c^2-2ac\,\, cos B-------(2)$

compare (1) abd (2)

if (1)=(2) then $cos B=\dfrac {1}{2},\,\, \,\,now\,\, we\,\,\,only\,\, have\,\, to \,\, prove : \angle B=60^o$

$using :\angle A+\angle C=2\angle B,\therefore\,\, 3\angle B=180^o \Longleftrightarrow \angle B=60^0$

and the proof is finished