In triangle $ABC$, $\angle ABC = 90^\circ$, and point $D$ lies on segment $BC$ such that $AD$ is an angle bisector. If $AB = 60$ and $BD = 40$, then find $AC$.

Answer 1

Let $E$ be the foot of the altitude from $D$ to $\overline{AB},$ as shown below.

[asy]
pair A,B,C,D,E;

A=(0,48); B=(0,0); C=(64,0); D=(12.8,0); E=(12.8,19.2);

draw(A--B--C--cycle);

draw(A--D);
draw(D--E);

pair[] ps={A,B,C,D,E};

dot(ps);

label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,S);
label("$E$",E,NE);

label("$60$",(A+B)/2,W);
label("$40$",(B+D)/2,S);
label("$x$",(A+D)/2,N);
[/asy]

It follows that, if $AD = x,$ then
\[\frac{60}{x} = \frac{64}{40}.\]Cross-multiplying gives $40x = 60 \cdot 64 = 3840,$ so $x = 96.$

By the Pythagorean Theorem on right triangle $ABD,$
\[AD = \sqrt{AB^2 - BD^2} = \sqrt{60^2 - 40^2} = \sqrt{4 \cdot 400} = 80.\]Then by the Angle Bisector Theorem, $CD = \frac{60 \cdot 40}{80} = 30,$ so $AC = AD + CD = 80 + 30 = \boxed{110}.$