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Question 82
The bisector of $$\angle B$$ in $$\triangle ABC$$ meets $$AC$$ at $$D$$. If $$AB = 10 cm, BC = 11 cm$$ and $$AC = 14 cm$$, then the length of $$AD$$ is:
Solution
by the property,
$$\frac{BC}{CD} = \frac{AB}{AD}$$
$$\frac{11}{AC - AD} = \frac{10}{AD}$$
$$\frac{11}{14 - AD} = \frac{10}{AD}$$
11AD = 140 - 10AD
21AD = 140
AD = $$\frac{20}{3} cm$$
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