Let $E$ be on $\overline{AB}$ such that $\overline{CE}$ is an angle bisector. Note that triangle $ABC$ has area
\[\frac{1}{2} \cdot 2 \cdot 2 \cdot \sin A = 2 \sin A,\]and triangle $ACE$ has area $\frac{1}{2} \cdot 2 \cdot 2 \cdot \sin \frac{A}{2} \cos \frac{A}{2} = 2 \sin \frac{A}{2} \cos \frac{A}{2} = \sin A.$
[asy]
unitsize (0.5 cm);
pair A, B, C, D, E;
A = (5,4);
B = (0,0);
C = (20,0);
D = interp(B,C,2/3);
E = interp(A,B,3/5);
draw(A--B--C--cycle);
draw(A--D);
draw(A--E);
draw(C--E);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NW);
[/asy]
Then
\[\frac{BD}{CD} = \frac{AB}{AC} \cdot \frac{\sin \angle BDA}{\sin \angle CDA} = \frac{2}{2} \cdot \frac{\sin \angle BDA}{\sin \angle CDA} = \frac{\sin \angle BDA}{\sin \angle CDA}.\]Since triangles $ABD$ and $ACD$ have the same height (namely, $AD$) from bases $BD$ and $CD,$ respectively,
\[\frac{[ABD]}{[ACD]} = \frac{BD}{CD}.\]Thus,
\[\frac{[ABD]}{[ACD]} = \frac{\sin \angle BDA}{\sin \angle CDA} = \frac{\sin \angle BDA}{\sin \angle BEA} = \frac{DB}{EB} = \frac{2}{2} = \boxed{1}.\]
In triangle $ABC$, $D$ lies on segment $\overline{BC}$ such that $\overline{AD}$ is an angle bisector. If $AB = 2$, $AC = 2$, and $BC = 19$, then find the ratio of the area of triangle $ABD$ to the area of triangle $ACD$. (Express your answer as a fraction in lowest terms.)
1 answer