In `DeltaABC,ADbotBCandAD^(2)=BD*CD` . Prove that `angleBAC=90^(@)` .

Answer

To prove that angle \( \angle BAC = 90^\circ \) in triangle \( \Delta ABC \) given that \( AD \bot BC \) and \( AD^2 = BD \cdot CD \), we can follow these steps:

### Step 1: Understand the Given Information
We are given a triangle \( \Delta ABC \) with a point \( D \) on line segment \( BC \) such that \( AD \) is perpendicular to \( BC \). This means that \( \angle ADB = 90^\circ \).

### Step 2: Apply the Pythagorean Theorem
In triangle \( ADB \):
- By the Pythagorean theorem, we have:
\[ AB^2 = AD^2 + BD^2 \quad \text{(1)} \]

In triangle \( ADC \):
- Again, by the Pythagorean theorem, we have:
\[ AC^2 = AD^2 + CD^2 \quad \text{(2)} \]

### Step 3: Combine the Equations
Now we can add equations (1) and (2):
\[ AB^2 + AC^2 = (AD^2 + BD^2) + (AD^2 + CD^2) \]
This simplifies to:
\[ AB^2 + AC^2 = 2AD^2 + BD^2 + CD^2 \quad \text{(3)} \]

### Step 4: Substitute the Given Condition
From the problem, we know that:
\[ AD^2 = BD \cdot CD \]
Substituting this into equation (3):
\[ AB^2 + AC^2 = 2(BD \cdot CD) + BD^2 + CD^2 \]

### Step 5: Simplify the Right Side
We can factor the right side:
\[ AB^2 + AC^2 = BD^2 + 2BD \cdot CD + CD^2 \]
This can be rewritten as:
\[ AB^2 + AC^2 = (BD + CD)^2 \quad \text{(4)} \]

### Step 6: Conclude the Proof
Now, from equation (4), we have:
\[ AB^2 + AC^2 = (BD + CD)^2 \]
This is the form of the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Thus, we conclude that:
\[ \angle BAC = 90^\circ \]

### Final Conclusion
We have proved that \( \angle BAC = 90^\circ \) as required.

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