Please help me with this trig(properties of triangles) problem?

[rumford]

If acosA = bcosB then show that the triangle is either an isosceles triangle or right angled triangle.

[eddis]

i) Applying sine law of triangles, a/sinA = b/sinB = c/sinC = 2R, where R is the circum radius,
a = 2R*sin(A) and b = 2R*sin(B)

ii) Hence, substituting this in the given equation,
2R*sin(A)cos(A) = 2R*sin(B)cos(B)
==> 2sin(A)cos(A) = 2sin(B)cos(B)

==> sin(2A) = sin(2B) [Application of multiple angle identity]

==> Either 2A = 2B or 2A = 180 - 2B, since sin(180 - 2B) = sin(2B)

When 2A = 2B, the angles are equal, so the triangle is isosceles.
When 2A = 180 - 2B, 2A + 2B = 180; ==> A+B = 90 deg; so the triangle is right triangle.

Thus either the triangle is isosceles or right triangle is proved.