ABC is a right triangle with angle B as the right angle. AD = 3 and DC = 12.

What are the lengths of AB and BC?
BD is the altitude to the hypotenuse AC.

If an altitude is drawn from the vertex with the right angle to the hypotenuse then the triangle is divided into two smaller triangles which are both similar to the original and therefore similar to each other.

The altitude is the geometric mean (mean proportional) of the two segments of the hypotenuse.
Each leg of the triangle is the mean proportional of the hypotenuse and the adjacent segment.
In equations,

Altitude^2 = 12 x 3
Altitude = √36 = 6

AB^2 = √(3 x 15)
AB = 3√5

BC^2 = √(12 x 15)
BC = 6√5

Given altitude drawn to the hypotenuse of the right triangle, length of the leg is equal to the geometric mean between the hypotenuse and segment of hypotenuse that is adjacent to the leg. So,

AB = sqrt(AC * AD)
= sqrt(15 * 3) = 3 sqrt(5)

BC = sqrt(AC * CD)
= sqrt(15 * 12) = 6 sqrt(5)

Start out by drawing this out. Here is a diagram I made:
https://mail.google.com/mail/?ui=2&ik=aa…

Note that triangles ABC and BCD are similar, so we have:
BC/12 = 15/BC.

Cross-multiplying yields:
BC^2 = 15 * 12 = 180 ==> BC = 6√5.

Then, since AC = AD + DC = 3 + 12 = 15, we see that, by applying the Pythagorean Theorem on triangle ABC:
AB^2 + BC^2 = AC^2.

With BC = 6√5 and AC = 15:
AB^2 + (6√5)^2 = 15^2 ==> AB = √[15^2 - (6√5)^2] = 3√5.

I hope this helps!