ABC is an isosceles triangle with AB = AC. A circle through B touching AC at the middle point Intersects AB at P. Then AP: AB is :

Options

A. `4:1`
B. `2:3`
C. `3:5`
D. `1:4 `

Answer

To solve the problem, we need to find the ratio \( AP : AB \) in the isosceles triangle \( ABC \) where \( AB = AC \) and a circle through point \( B \) touches line \( AC \) at its midpoint.

### Step-by-Step Solution:

1. **Identify the Triangle and Points**:
- Let \( A \) be the vertex of the triangle, and \( B \) and \( C \) be the other two vertices such that \( AB = AC \).
- Let \( M \) be the midpoint of \( AC \).

2. **Circle Touching AC**:
- A circle passes through point \( B \) and touches line \( AC \) at point \( M \).

3. **Intersecting AB**:
- The circle intersects line \( AB \) at point \( P \).

4. **Using the Power of a Point Theorem**:
- According to the Power of a Point theorem, we have:
\[ AP \cdot AB = AM^2 \]
- Since \( M \) is the midpoint of \( AC \), we can express \( AM \) in terms of \( AB \):
\[ AM = \frac{AC}{2} = \frac{AB}{2} \]

5. **Substituting into the Equation**:
- Substitute \( AM \) into the equation:
\[ AP \cdot AB = \left(\frac{AB}{2}\right)^2 \]
- Simplifying this gives:
\[ AP \cdot AB = \frac{AB^2}{4} \]

6. **Solving for AP**:
- Now, divide both sides by \( AB \) (assuming \( AB \neq 0 \)):
\[ AP = \frac{AB^2}{4 \cdot AB} = \frac{AB}{4} \]

7. **Finding the Ratio**:
- Now we can find the ratio \( AP : AB \):
\[ AP : AB = \frac{AB}{4} : AB = 1 : 4 \]

Thus, the final answer is:
\[ AP : AB = 1 : 4 \]