In a triangle ABC, if `(a+b+c)(a+b-c)(b+c-a)(c+a-b)=(8a^2b^2c^2)/(a^2+b^2+c^2)` then the triangle is

Options

A. isosceles
B. right angled
C. equilateral
D. obtuse angled

Answer

To solve the problem, we need to analyze the given equation and determine the type of triangle it represents. Let's break down the solution step by step.

### Step 1: Understand the Given Equation
The equation provided is:
\[ (a+b+c)(a+b-c)(b+c-a)(c+a-b) = \frac{8a^2b^2c^2}{a^2+b^2+c^2} \]
This equation involves the sides of triangle ABC, denoted as \(a\), \(b\), and \(c\).

### Step 2: Use the Semi-perimeter
Let the semi-perimeter \(s\) of the triangle be defined as:
\[ s = \frac{a+b+c}{2} \]
Then, we can express the terms \(a+b-c\), \(b+c-a\), and \(c+a-b\) in terms of \(s\):
\[ a+b-c = 2s - 2c, \quad b+c-a = 2s - 2a, \quad c+a-b = 2s - 2b \]

### Step 3: Substitute into the Equation
Substituting these expressions into the original equation gives:
\[ (2s)(2s - 2c)(2s - 2a)(2s - 2b) = \frac{8a^2b^2c^2}{a^2 + b^2 + c^2} \]
This simplifies to:
\[ 16s(s-a)(s-b)(s-c) = \frac{8a^2b^2c^2}{a^2 + b^2 + c^2} \]

### Step 4: Relate to Area of the Triangle
The area \(K\) of triangle ABC can be expressed using Heron's formula:
\[ K = \sqrt{s(s-a)(s-b)(s-c)} \]
Thus, we can express \(s(s-a)(s-b)(s-c)\) in terms of the area:
\[ s(s-a)(s-b)(s-c) = \frac{K^2}{s} \]
Substituting this into our equation gives:
\[ 16 \cdot \frac{K^2}{s} = \frac{8a^2b^2c^2}{a^2 + b^2 + c^2} \]

### Step 5: Rearranging the Equation
Cross-multiplying leads to:
\[ 16K^2(a^2 + b^2 + c^2) = 8a^2b^2c^2 \]
Dividing both sides by 8 gives:
\[ 2K^2(a^2 + b^2 + c^2) = a^2b^2c^2 \]

### Step 6: Analyze the Result
According to the sine rule, we can express the sides in terms of the circumradius \(R\):
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]
Substituting these into the equation will lead to a relationship involving \(R\) and the angles \(A\), \(B\), and \(C\).

### Step 7: Conclusion
From the derived equation, we find that:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 2 \]
This condition is satisfied when one of the angles \(A\), \(B\), or \(C\) is \(90^\circ\), indicating that the triangle is a right triangle.

Thus, the triangle ABC is a **right-angled triangle**.