In \(\triangle ABC\), if \(a\cos A=b\cos B\) then prove that the triangle is either a right angled or an isosceles triangle.

Answer: Proved (\(A=B\) or \(C=90^\circ\)).

By the sine rule, \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=k\), so \(a=k\sin A\), \(b=k\sin B\).

Given \(a\cos A=b\cos B\): \(k\sin A\cos A=k\sin B\cos B\Rightarrow 2\sin A\cos A=2\sin B\cos B\Rightarrow\sin2A=\sin2B\).

\(\therefore\sin2A-\sin2B=0\Rightarrow2\cos(A+B)\sin(A-B)=0\).

So either \(\sin(A-B)=0\Rightarrow A=B\) (triangle is isosceles), or \(\cos(A+B)=0\Rightarrow A+B=\dfrac{\pi}{2}\Rightarrow C=\pi-(A+B)=\dfrac{\pi}{2}\) (triangle is right angled).

Hence the triangle is either right angled or isosceles.