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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 & Solution
Correct Answer:
Option
D
According to question,
Let AB = AC = 2x
∵ AQ = QC = x
∴ AB is a secant
∴ AP × AB = AQ2
AP × 2x = x2
$$\eqalign{ & AP = \frac{x}{2} \cr & \frac{{AP}}{{AB}} = \frac{x}{{2 \times 2x}} = \frac{1}{4} \cr & \frac{{AP}}{{AB}} = \frac{1}{4} \cr & AP:AB = 1:4 \cr} $$
Let AB = AC = 2x
∵ AQ = QC = x
∴ AB is a secant
∴ AP × AB = AQ2
AP × 2x = x2
$$\eqalign{ & AP = \frac{x}{2} \cr & \frac{{AP}}{{AB}} = \frac{x}{{2 \times 2x}} = \frac{1}{4} \cr & \frac{{AP}}{{AB}} = \frac{1}{4} \cr & AP:AB = 1:4 \cr} $$
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