In a triangle ABC, prove that `(cot(A/2)+cot(B/2)+cot(C/2))/(cotA+cotB+cot(C))=((a+b+c)^2)/(a^2+b^2+c^2)`

Answer

To prove the equation

\[ \frac{\cot(A/2) + \cot(B/2) + \cot(C/2)}{\cot A + \cot B + \cot C} = \frac{(a+b+c)^2}{a^2 + b^2 + c^2} \]

we will follow these steps:

### Step 1: Express cot(A/2), cot(B/2), and cot(C/2) using known formulas.
The cotangent of half-angles can be expressed as:

\[ \cot\left(\frac{A}{2}\right) = \frac{s(s-a)}{r} \]
\[ \cot\left(\frac{B}{2}\right) = \frac{s(s-b)}{r} \]
\[ \cot\left(\frac{C}{2}\right) = \frac{s(s-c)}{r} \]

where \( s = \frac{a+b+c}{2} \) is the semi-perimeter and \( r \) is the inradius.

### Step 2: Substitute the cotangent values into the left-hand side (LHS).
Now substituting these into the LHS:

\[ \cot\left(\frac{A}{2}\right) + \cot\left(\frac{B}{2}\right) + \cot\left(\frac{C}{2}\right) = \frac{s(s-a)}{r} + \frac{s(s-b)}{r} + \frac{s(s-c)}{r} \]

Combining these gives:

\[ \frac{s[(s-a) + (s-b) + (s-c)]}{r} = \frac{s[3s - (a+b+c)]}{r} = \frac{s[3s - 2s]}{r} = \frac{s^2}{r} \]

### Step 3: Express cot A, cot B, and cot C using known formulas.
The cotangent of the angles can be expressed as:

\[ \cot A = \frac{b^2 + c^2 - a^2}{4\Delta}, \quad \cot B = \frac{a^2 + c^2 - b^2}{4\Delta}, \quad \cot C = \frac{a^2 + b^2 - c^2}{4\Delta} \]

where \( \Delta \) is the area of triangle ABC.

### Step 4: Substitute the cotangent values into the right-hand side (RHS).
Now substituting these into the denominator of the LHS:

\[ \cot A + \cot B + \cot C = \frac{(b^2+c^2-a^2) + (a^2+c^2-b^2) + (a^2+b^2-c^2)}{4\Delta} \]

This simplifies to:

\[ \frac{2(a^2 + b^2 + c^2)}{4\Delta} = \frac{a^2 + b^2 + c^2}{2\Delta} \]

### Step 5: Combine LHS and RHS.
Now we can express the LHS as:

\[ \frac{\frac{s^2}{r}}{\frac{a^2 + b^2 + c^2}{2\Delta}} = \frac{2s^2 \Delta}{r(a^2 + b^2 + c^2)} \]

### Step 6: Use the relationship between area, semi-perimeter, and inradius.
From the relationship \( \Delta = rs \), we substitute \( r = \frac{\Delta}{s} \):

\[ \frac{2s^2 \Delta}{\Delta(a^2 + b^2 + c^2)} = \frac{2s^2}{a^2 + b^2 + c^2} \]

### Step 7: Substitute \( s = \frac{a+b+c}{2} \).
Substituting \( s \):

\[ \frac{2\left(\frac{a+b+c}{2}\right)^2}{a^2 + b^2 + c^2} = \frac{(a+b+c)^2}{a^2 + b^2 + c^2} \]

### Conclusion:
Thus, we have shown that:

\[ \frac{\cot(A/2) + \cot(B/2) + \cot(C/2)}{\cot A + \cot B + \cot C} = \frac{(a+b+c)^2}{a^2 + b^2 + c^2} \]

Hence, the statement is proved.

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