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zloy xaker [14]
3 years ago
13

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.)'

Business
1 answer:
Hunter-Best [27]3 years ago
6 0

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)

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