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 Circle**:
- Let \( AB = AC = a \).
- Let \( M \) be the midpoint of \( AC \).
- The circle passes through \( B \) and touches \( AC \) at point \( M \).

2. **Use the Power of a Point Theorem**:
- According to the Power of a Point theorem, if a circle touches a line at a point, the power of the point (in this case, point \( A \)) with respect to the circle can be expressed as:
\[ AP \times AB = AM^2 \]
- Here, \( AM \) is the distance from point \( A \) to point \( M \).

3. **Calculate \( AM \)**:
- Since \( M \) is the midpoint of \( AC \), we have:
\[ AM = \frac{AC}{2} = \frac{a}{2} \]
- Therefore, \( AM^2 = \left(\frac{a}{2}\right)^2 = \frac{a^2}{4} \).

4. **Substitute into the Power of a Point Equation**:
- Now substituting \( AM^2 \) into the equation gives:
\[ AP \times AB = \frac{a^2}{4} \]

5. **Express \( AP \) in terms of \( AB \)**:
- Since \( AB = a \), we can substitute \( AB \) into the equation:
\[ AP \times a = \frac{a^2}{4} \]
- Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ AP = \frac{a}{4} \]

6. **Find the Ratio \( AP : AB \)**:
- Now we can express the ratio \( AP : AB \):
\[ AP : AB = \frac{a}{4} : a \]
- Simplifying this gives:
\[ AP : AB = 1 : 4 \]

### Final Answer:
Thus, the required ratio \( AP : AB \) is \( 1 : 4 \).