ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. The ratio between the areas of `Delta ABE and Delta ACD` is

Options

A. `sqrt(2) : 1`
B. `1 : 2`
C. `2 : 1`
D. `sqrt(2) : 1`

Answer

To find the ratio between the areas of triangles ABE and ACD, we can follow these steps:

### Step 1: Understand the Properties of Triangle ABC
Given that triangle ABC is an isosceles triangle right-angled at B, we know that:
- AB = BC (since it's isosceles)
- The angle at B is 90 degrees.

### Step 2: Use the Pythagorean Theorem
In triangle ABC, we can apply the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Since AB = BC, we can substitute:
\[ AC^2 = AB^2 + AB^2 = 2AB^2 \]
Thus, we have:
\[ AC^2 = 2AB^2 \]
This will be useful later.

### Step 3: Area of Similar Triangles
The areas of similar triangles are proportional to the squares of their corresponding sides. Therefore, we can express the ratio of the areas of triangles ABE and ACD as:
\[ \frac{\text{Area of } \triangle ABE}{\text{Area of } \triangle ACD} = \left(\frac{AB}{AC}\right)^2 \]

### Step 4: Substitute the Values
From our previous step, we know:
\[ AC^2 = 2AB^2 \]
So, we can find the ratio:
\[ \frac{AB}{AC} = \frac{AB}{\sqrt{2AB^2}} = \frac{AB}{AB\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Calculate the Ratio of Areas
Now, substituting this into our area ratio:
\[ \frac{\text{Area of } \triangle ABE}{\text{Area of } \triangle ACD} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

### Final Answer
Thus, the ratio of the areas of triangles ABE and ACD is:
\[ \frac{\text{Area of } \triangle ABE}{\text{Area of } \triangle ACD} = \frac{1}{2} \]

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