Question

Solve:

$\frac{c-bcosA}{b-ccosA}=\frac{cosA}{cosC}$

Answer 1

Given, $\frac{c-b\cos A}{b-c\cos A}=\frac{\cos A}{\cos C}$

$\Rightarrow \frac{c-\left(\frac{b^2+c^2-a^2}{2c}\right)}{b-\left(\frac{b^2+c^2-a^2}{2b}\right)}=\frac{\frac{b^2+c^2-a^2}{2bc}}{\frac{b^2+c^2-a^2}{2ab}}$ (cosine rule)

$\Rightarrow \frac{\frac{a^2+c^2-b^2}{2c}}{\frac{a^2+b^2-c^2}{2b}}=\frac{\frac{b^2+c^2-a^2}{2bc}}{\frac{a^2+b^2-c^2}{2ab}}$

$\Rightarrow b{a}^2+b{c}^2-b^3=a{b}^2+a{c}^2-a^3$

$\Rightarrow ab(a-b)+c^2(b-a)+a^3-b^3=0$

$\Rightarrow (a-b)[ab+(-c^2)+a^2+b^2+ab]=0$

$\Rightarrow a=b$.....................................(1)

or $a^2+b^2+2ab=c^2$

$(a+b{)}^2=c^2[\because a+b+c\ne 0]$

$a+b=c$................................(2)