Prove the following equation

[Albert]

$\triangle ABC (with \,\,side \,\,length \,\, a,b,c), \,\, given :\angle A+\angle C =2 \angle B $

prove :

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

 

[Opalg]

Spoiler

\begin{array}{rcl}\dfrac {1}{a+b}+\dfrac{1}{b+c}=\dfrac {3}{a+b+c} & \Longleftrightarrow & (a+2b+c)(a+b+c) = 3(a+b)(a+c) \\ & \Longleftrightarrow & a^2+b^2+c^2 + 3ab + 3bc + 2ac = 3b^2 + 3ab + 3bc + 3ac \\ & \Longleftrightarrow & b^2 = a^2 + c^2 - ac. \end{array}
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.

 

[Jester]

Spoiler

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

 

[Opalg]

Spoiler

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

 

[Albert]

from Opalg's solution :
$b^2 = a^2 + c^2 - ac.$--------(1)
as jester said "Couldn't you use the law of Cosine with [FONT=MathJax_Main]∠[/FONT][FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]60[/FONT][FONT=MathJax_Main]∘[/FONT] to get to the same answer quicker?"
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