In `Delta ABC AD ` divides BC in 1 : 3 find the ratio of triangle ADB and ADC ?

Answer

To find the ratio of the areas of triangles ADB and ADC, we can follow these steps:

### Step 1: Understand the problem
We have triangle ABC with point D on line segment BC such that AD divides BC in the ratio 1:3. This means that BD:DC = 1:3.

### Step 2: Assign lengths to segments
Let the length of segment BD be \( x \). Since BD:DC = 1:3, the length of segment DC will be \( 3x \). Therefore, the total length of segment BC is:
\[ BC = BD + DC = x + 3x = 4x \]

### Step 3: Calculate the areas of triangles ADB and ADC
The area of a triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

- For triangle ADB:
- Base = BD = \( x \)
- Height = h (the perpendicular distance from A to line BC)

Thus, the area of triangle ADB is:
\[ \text{Area}_{ADB} = \frac{1}{2} \times x \times h \]

- For triangle ADC:
- Base = DC = \( 3x \)
- Height = h (the height remains the same since it is the same vertex A)

Thus, the area of triangle ADC is:
\[ \text{Area}_{ADC} = \frac{1}{2} \times 3x \times h \]

### Step 4: Find the ratio of the areas
Now, we can find the ratio of the areas of triangles ADB and ADC:
\[ \text{Ratio} = \frac{\text{Area}_{ADB}}{\text{Area}_{ADC}} = \frac{\frac{1}{2} \times x \times h}{\frac{1}{2} \times 3x \times h} \]

### Step 5: Simplify the ratio
The \(\frac{1}{2}\) and \(h\) cancel out:
\[ \text{Ratio} = \frac{x}{3x} = \frac{1}{3} \]

Thus, the ratio of the areas of triangles ADB and ADC is:
\[ \text{Ratio of Area}_{ADB} : \text{Area}_{ADC} = 1 : 3 \]

### Final Answer:
The ratio of the areas of triangles ADB and ADC is \( 1 : 3 \).

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