Question 61
In $$\triangle ABC, AB = AC$$. A circle drawn through B touches AC at D and intersect AB at P. If D is the mid point of AC and AP 2.5 cm, then AB is equal to:
Solution
Given D is midpoint of AC so,
AD = $$\frac{AC}{2}$$
But also given AC = AB
AD = $$\frac{AB}{2}$$ ----(1)
AD is a tangent and APB is a secant. So the tangent secant theorem can be applied,
$$AD^2 = AP \times AB$$
$$(\frac{AB}{2})^2 = 2.5 \times AB$$
$$\frac{AB^2}{4} = 2.5 \times AB$$
AB = 10 cm
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