In triangle $ABC$, $D$ lies on segment $\overline{BC}$ such that $\overline{AD}$ is an angle bisector. If $AB = 2$, $AC = 2$, and $BC = 3$, 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.)

Answer 1

From the Angle Bisector Theorem, we have $\displaystyle\frac{BD}{CD} = \frac{AB}{AC} = 1$. We proceed by dropping an perpendicular from $A$ to $\overline{BC}$, as shown below: [asy]
pair A,B,C,D,EE;
A = (0,0);
B = (1,0);
C= (2.5,0);
EE = (1.5,0);
D = (2,0);
draw(A--B--C--cycle);
draw(A--D);
draw(A--EE);
label("$A$",A,S);
label("$B$",B,S);
label("$D$",D,S);
label("$C$",C,S);
label("$E$",EE,S);
label("$2$",(A+EE)/2,S);
label("$2$",(EE+C)/2,S);
label("$3$",(B+C)/2,S);
[/asy] Therefore, by SAS, we have $\triangle ACD \cong \triangle ABD$, from which it follows that $[ABD] = [ACD]$ and the ratio of the areas of the two triangles is $[ABD]:[ACD] = \boxed{1:1}$.