Given:
In triangle ABC,
2sinAcosB = SinA
2sinBcosC = SinB
2sinCcosA = SinC

To prove:
Triangle ABC is a right-angled triangle.

Proof:
Let's assume that triangle ABC is not a right-angled triangle. In that case, we have three possibilities:

Case 1: Triangle ABC is an acute-angled triangle.
In an acute-angled triangle, all angles are less than 90 degrees. Since the sine of an acute angle is always positive, we can conclude that sinA, sinB, and sinC are all positive. Similarly, cosB, cosC, and cosA are also positive in an acute-angled triangle.

Now, using the given equations:
2sinAcosB = sinA
2sinBcosC = sinB
2sinCcosA = sinC

Since sinA, sinB, and sinC are positive, we can divide each equation by sinA, sinB, and sinC respectively to get:
2cosB = 1
2cosC = 1
2cosA = 1

But this leads to a contradiction because the cosine of an acute angle is always less than 1. Therefore, triangle ABC cannot be an acute-angled triangle.

Case 2: Triangle ABC is an obtuse-angled triangle.
In an obtuse-angled triangle, one angle is greater than 90 degrees. Without loss of generality, let's assume that angle A is obtuse. In this case, sinA is negative and cosA is positive.

Using the given equations:
2sinAcosB = sinA
2sinBcosC = sinB
2sinCcosA = sinC

Since sinA is negative, we can divide the first equation by sinA to get:
2cosB = -1

But this leads to a contradiction because the cosine of an angle cannot be negative. Therefore, triangle ABC cannot be an obtuse-angled triangle.

Case 3: Triangle ABC is a right-angled triangle.
In a right-angled triangle, one angle is exactly 90 degrees. Without loss of generality, let's assume that angle A is 90 degrees. In this case, sinA is 1 and cosA is 0.

Using the given equations:
2sinAcosB = sinA
2sinBcosC = sinB
2sinCcosA = sinC

Substituting sinA = 1 and cosA = 0 in the third equation, we get:
2sinCcosA = sinC
2sinC(0) = sinC
0 = sinC

This implies that angle C is also 90 degrees. Therefore, triangle ABC is a right-angled triangle.

Conclusion:
From the above analysis, we can conclude that if the given equations hold true, then triangle ABC must be a right-angled triangle.

Equilateral triangle