Find the Sum of Reciprocals

Geometry Level 3

In $\triangle ABC$ , sides $BC$ , $AC$ , and $AB$ have lengths $a$ , $b$ , and $c$ respectively. The angle bisector of $\angle BAC$ intersects $BC$ at $D$ . If $\cos (\angle BAD) = \dfrac 13$ and $AD = 6$ , find $\dfrac{1}{b} + \dfrac{1}{c}$ .

1/6 1/9 2/5 2/3

1 solution

Jam M
Feb 6, 2020

Let d = AD and $\alpha = \angle BAD = \angle CAD$ . $[ABC] = [BAD] + [CAD]$ $\dfrac{bc}{2} \sin(2\alpha) = \dfrac{cd}{2} \sin(\alpha) + \dfrac{bd}{2} \sin(\alpha)$ $bc \cos(\alpha) = \dfrac{1}{2}d(c+b)$ $\dfrac{b+c}{bc} = \dfrac{2\cos(\alpha)}{d}$ $\dfrac{1}{b} + \dfrac{1}{c} = \dfrac{2(1/3)}{6} = \dfrac{1}{9}$

Find the Sum of Reciprocals

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