91Ó°ÊÓ

In any triangle, the sine rule states that \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) Given: \(a \sin A = b \sin B\), we will rewrite the equation using sine rule: \(\frac{a^2}{a \sin A} = \frac{b^2}{b \sin B}\)

Now we can simplify the equation: \[\frac{a}{\sin A} = \frac{b}{\sin B}\] Since \(\frac{a}{\sin A} = \frac{b}{\sin B}\), using the sine rule, we get: \[\frac{c}{\sin C} = \frac{a}{\sin A} = \frac{b}{\sin B}\]

There are two possibilities that would make the triangle isosceles: 1. Angle A = Angle B 2. Angle A = Angle C or Angle B = Angle C

Using the given equation \(a \sin A = b \sin B\) and sine rule, let's analyze the possibilities. 1. If angle A = angle B, then \(\sin A = \sin B\) and using the sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\). This possibility holds true since the conditions are already met. 2. If angle A = angle C or angle B = angle C, then consider when A = C. We will have \(\sin A = \sin C\). From sine rule, we have \(\frac{a}{\sin A} = \frac{c}{\sin C}\). Substituting \(\sin A\) with \(\sin C\), we get \(\frac{a}{\sin A} = \frac{c}{\sin A}\), which yields \(a=c\). The triangle becomes isosceles with A = C. Similarly, if B = C, we would have \(\sin B = \sin C\) and then \(\frac{b}{\sin B} = \frac{c}{\sin C}\). Substituting \(\sin B\) with \(\sin C\), we get \(\frac{b}{\sin B} = \frac{c}{\sin B}\), which yields \(b=c\). The triangle becomes isosceles with B = C.

Therefore, based on the given condition \(a \sin A = b \sin B\), it is proven that the triangle ABC is isosceles, either angle A = angle B, angle A = angle C or angle B = angle C.