# Question #f3da8

Feb 16, 2017

$A D = \frac{1}{3}$

#### Explanation:

As shown in the diagram, $A E \mathmr{and} C D$ are the two altitudes.

$\Delta B D C$ is a right-angled triangle.
Given $B E = 6 , E C = 8 , \mathmr{and} B D = 9$,
$\implies B C = B E + E C = 6 + 8 = 14$
Let $\angle D B C = x$
$\cos x = \frac{B D}{B C} = \frac{9}{14}$

$\Delta A B E$ is also a right triangle, right-angled at $E$,
$\implies A B \cos x = 6$
$\implies A B = \frac{6}{\cos} x = \frac{6}{\frac{9}{14}} = 6 \times \frac{14}{9} = \frac{28}{3}$

$A D = A B - B D = \frac{28}{3} - 9 = \frac{1}{3}$