Question

In $△ABC$ is $\frac{\cos A}{a}=\frac{\cos B}{b}$

Show that it is an isosceles triangle.(With the standard notations.)

Answer 1

$\frac{\cos A}{a}=\frac{\cos B}{b}$

$\frac{b^2+c^2-a^2}{2bc\times a}=\frac{a^2+c^2-b^2}{2ac\times b}$

$b^2+c^2-a^2=a^2+c^2-b^2$

$b^2-a^2=a^2-b^2$

$a^2=b^2$

$a=b$

Therefore, it shows $\Delta ABC$ is an isosceles triangle.