ABC is an isosceles trianlge with AB=AC. A circle through B touches side AC at its middle point D and intersects side AB in point P. Show that `AB=4xxAP`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Draw the Isosceles Triangle
Draw triangle ABC such that AB = AC. Mark the midpoint of AC as point D.

### Step 2: Draw the Circle
Draw a circle that passes through point B and touches side AC at point D.

### Step 3: Identify the Tangent and Secant
Recognize that line segment AD is the tangent to the circle at point D, and line segment AB is the secant that intersects the circle at point P.

### Step 4: Apply the Tangent-Secant Theorem
According to the tangent-secant theorem, we have:
\[ AD^2 = AP \cdot AB \]
where:
- AD is the length of the tangent segment,
- AP is the length of the exterior portion of the secant,
- AB is the length of the whole secant.

### Step 5: Express AD in Terms of AB
Since D is the midpoint of AC, we have:
\[ AD = \frac{AC}{2} = \frac{AB}{2} \]
because AB = AC in an isosceles triangle.

### Step 6: Substitute AD in the Equation
Substituting AD in the tangent-secant theorem gives us:
\[ \left(\frac{AB}{2}\right)^2 = AP \cdot AB \]

### Step 7: Simplify the Equation
This simplifies to:
\[ \frac{AB^2}{4} = AP \cdot AB \]

### Step 8: Solve for AB
Dividing both sides by AB (assuming AB ≠ 0) gives:
\[ \frac{AB}{4} = AP \]
Thus, we can express AB as:
\[ AB = 4 \cdot AP \]

### Conclusion
We have shown that:
\[ AB = 4 \cdot AP \]

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