Question 59
In $$\triangle ABC,AB=7cm,BC=10cm,$$ and $$AC = 8 cm$$. If $$AD$$ is the angle bisector of $$\angle BAC$$, where $$D$$ is a point on $$BC$$, then $$BD$$ is equal to:
Solution
Let the BD = x,
DC = 10 - x
By angle bisector theorem,
$$\frac{BD}{CD} = \frac{AB}{AC}$$
$$\frac{x}{10 - x} = \frac{7}{8}$$
8x = 70 - 7x
15x = 70
x = 14/3 cm
$$BD$$ = $$\frac{14}{3}$$
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