Question 134

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

Here, AM is tangent to the circle and APB is secant to the circle. Thus by tangent theorem :

=> $$AM^2 = AP \times AB$$

Also, $$AM = \frac{AC}{2} = \frac{AB}{2}$$

=> $$(\frac{AB}{2})^2 = AP \times AB$$

=> $$\frac{AP}{AB} = \frac{1}{4}$$

=> Required ratio = 1 : 4

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