Question

In triangle $ABC$, $AB = 16$, $AC = 24$, $BC = 19$, and $AD$ is an angle bisector. Find the ratio of the area of triangle $ABD$ to the area of triangle $ACD$. (Express your answer as a fraction in lowest terms.)'

Answer 1

Answer:

$DC=\frac{57}{5}$

Explanation:

We have triagle ABC bisected by angle AD, like is shown in the image.

A bisector angle is a line which cuts an angle into two equal halves.

So:

BC = BD + DC ∴ 19  =BD + DC ⇒ BD = 19 - DC  (1) because D cuts BC in 2 parts.

By angle bisector theorem: the ratio of any 2 sides of a triagle is equal to the ratio of the lenths formed on its third side by the anfle bisectorof the anfle formed by those 2 sides:

$\frac{AB}{BD} = \frac{AC}{DC}$ (2) read: side AB is BD as AC is to DC

Clear DC:

$DC = \frac{AC*BD}{AB}$ (3)

we know: AC = 24, BD = 19 - DC (from (1)) and AB = 16

replacing all the values:

$DC = \frac{24 ( 19 - DC)}{16}$

$DC=\frac{57}{5}$ (4)

for (4) in (1)

BD = 19 - $\frac{57}{5}$=$\frac{38}{5}$ (5)