In `DeltaABC,AD` is the bisector of `angleA` . Then , `(ar(DeltaABD))/(ar(DeltaACD))=`

Options

A. `(AB^(2))/(AC^(2))`
B. `(AB)/(AC)`
C. `(BM)/(CM)`
D. None of these

Answer

To find the ratio of the areas of triangles ABD and ACD in triangle ABC, where AD is the angle bisector of angle A, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have triangle ABC with AD as the angle bisector of angle A. We need to find the ratio of the areas of triangles ABD and ACD.

2. **Using the Angle Bisector Theorem**: According to the angle bisector theorem, the ratio of the lengths of the two segments created by the angle bisector on the opposite side is equal to the ratio of the other two sides of the triangle. Therefore, we have:
\[ \frac{AB}{AC} = \frac{BD}{DC} \]

3. **Area of Triangle Formula**: The area of a triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
For triangle ABD, the base is BD and the height is the perpendicular dropped from A to line BD. For triangle ACD, the base is DC and the height is the same perpendicular from A.

4. **Calculating Areas**:
- Area of triangle ABD:
\[ \text{Area}_{ABD} = \frac{1}{2} \times BD \times h \]
- Area of triangle ACD:
\[ \text{Area}_{ACD} = \frac{1}{2} \times DC \times h \]

5. **Finding the Ratio of Areas**:
Now we can find the ratio of the areas of triangles ABD and ACD:
\[ \frac{\text{Area}_{ABD}}{\text{Area}_{ACD}} = \frac{\frac{1}{2} \times BD \times h}{\frac{1}{2} \times DC \times h} \]
Here, the \(\frac{1}{2}\) and \(h\) cancel out:
\[ \frac{\text{Area}_{ABD}}{\text{Area}_{ACD}} = \frac{BD}{DC} \]

6. **Substituting from the Angle Bisector Theorem**:
From the angle bisector theorem, we know that:
\[ \frac{BD}{DC} = \frac{AB}{AC} \]
Therefore, we can conclude:
\[ \frac{\text{Area}_{ABD}}{\text{Area}_{ACD}} = \frac{AB}{AC} \]

### Final Result:
\[ \frac{\text{Area}_{ABD}}{\text{Area}_{ACD}} = \frac{AB}{AC} \]