In an acute angled triangle ABC, AD is the median in it. then :
` AD^(2)` =

Options

A. `(AB^(2))/3 + (AC^(2))/2 - (BC^(2))/4`
B. `(AB^(2))/2 + (AC^(2))/2 - (BC^(2))/4`
C. `(AB^(2))/2 + (AC^(2))/3 - (BC^(2))/4`
D. None

Answer

To prove that in an acute-angled triangle ABC, where AD is the median, the formula for \( AD^2 \) is given by:

\[ AD^2 = \frac{AB^2}{2} + \frac{AC^2}{2} - \frac{BC^2}{4} \]

we will follow these steps:

### Step 1: Set Up the Triangle
Let triangle ABC be an acute-angled triangle with vertices A, B, and C. Let D be the midpoint of side BC. By definition of a median, AD connects vertex A to the midpoint D of side BC.

### Step 2: Draw the Altitude
Draw a perpendicular from point A to line BC, and let the foot of the perpendicular be point E. This creates two right triangles, ABE and ACE.

### Step 3: Apply Pythagorean Theorem
Using the Pythagorean theorem in triangles ABE, ADE, and ACE, we can write:

1. For triangle ABE:
\[ AB^2 = AE^2 + BE^2 \quad \text{(1)} \]

2. For triangle ADE:
\[ AD^2 = AE^2 + DE^2 \quad \text{(2)} \]

3. For triangle ACE:
\[ AC^2 = AE^2 + CE^2 \quad \text{(3)} \]

### Step 4: Express BE and CE in terms of BD and CD
Since D is the midpoint of BC, we have:
\[ BE = BD \quad \text{and} \quad CE = CD \]
Thus, we can express BE and CE in terms of BD and CD.

### Step 5: Add Equations
Now, we will add equations (1) and (3):
\[ AB^2 + AC^2 = (AE^2 + BE^2) + (AE^2 + CE^2) \]
This simplifies to:
\[ AB^2 + AC^2 = 2AE^2 + BE^2 + CE^2 \]

### Step 6: Substitute BE and CE
Since \( BE = BD \) and \( CE = CD \), and \( BD = CD = \frac{BC}{2} \):
\[ BE^2 + CE^2 = BD^2 + CD^2 = 2 \left( \frac{BC}{2} \right)^2 = \frac{BC^2}{2} \]

### Step 7: Substitute Back
Now substituting back into the equation:
\[ AB^2 + AC^2 = 2AE^2 + \frac{BC^2}{2} \]

### Step 8: Rearranging
Rearranging gives:
\[ 2AE^2 = AB^2 + AC^2 - \frac{BC^2}{2} \]

### Step 9: Relate AE and AD
From equation (2), we have:
\[ AD^2 = AE^2 + DE^2 \]
Since DE is half of BC (because D is the midpoint), we can express DE in terms of BC:
\[ DE = \frac{BC}{2} \]
Thus:
\[ DE^2 = \left(\frac{BC}{2}\right)^2 = \frac{BC^2}{4} \]

### Step 10: Substitute DE^2 into AD^2
Substituting DE^2 in the expression for AD^2:
\[ AD^2 = AE^2 + \frac{BC^2}{4} \]

### Step 11: Combine Equations
Now substitute \( AE^2 \) from the rearranged equation:
\[ AD^2 = \frac{1}{2} \left( AB^2 + AC^2 - \frac{BC^2}{2} \right) + \frac{BC^2}{4} \]

### Step 12: Final Simplification
This simplifies to:
\[ AD^2 = \frac{AB^2}{2} + \frac{AC^2}{2} - \frac{BC^2}{4} \]

Thus, we have proved that:
\[ AD^2 = \frac{AB^2}{2} + \frac{AC^2}{2} - \frac{BC^2}{4} \]