How to find the measure of BC in this triangle?

Question

In this right triangle, AB = 1, BD= 1/2, and AD is a bisector. How do I find CD?

I tried finding AD with pythagora's theorem. Which would be square root of 3 over 2. Which should be equal to CD:

But it seems my logic is wrong. Can anyone help me?

Answer 1

Let $\theta:=\angle BAD$. Then angle $\angle BAC=2\theta$ so $$\frac {BC}{AB}=\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}.$$ This gives you $BC$, since you know $\tan\theta$ and you know $AB$.

(BTW, your calculation of $AD$ is incorrect.)

Answer 2

Suppose $CD=x$, then by Pythagoras's theorem, $AC=\sqrt{1+(x+\frac{1}{2})^2}$. Denote the intersection point on AC in your picture by E. Then $CE=AC-1$. $DE=DB=\frac{1}{2}$. And then use Pythagoras's theorem on $\Delta CDE$. Solve the equation.