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:

Solution (By Examveda Team)

According to question,

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