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:

Answer: Option D

According to question,

Let AB = AC = 2x
∵ AQ = QC = x
∴ AB is a secant
∴ AP × AB = AQ²
AP × 2x = x²
$$\begin{array}{rl}& AP=\frac{x}{2}\\ & \frac{AP}{AB}=\frac{x}{2\times 2x}=\frac{1}{4}\\ & \frac{AP}{AB}=\frac{1}{4}\\ & AP:AB=1:4\end{array}$$