In the diagram, $\triangle A B C$ is right-angled at $C$. Point $D$ is on $A C$ so that $\angle A B C=2 \angle D B C$. If $D C=1$ and $B D=3$, determine the length of $A D$.

![](https://cdn.mathpix.com/cropped/2024_04_30_fa78072b1e75930324fdg-3.jpg?height=328&width=371&top_left_y=1170&top_left_x=931)
Suppose that $\angle D B C=\theta$. Then $\angle A B C=2 \theta$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-12.jpg?height=336&width=379&top_left_y=1499&top_left_x=927)

Now $A D=A C-D C=A C-1$.

Also, $\tan (\angle A B C)=\tan 2 \theta=\frac{A C}{B C}$, so $A C=B C \tan 2 \theta$.

In $\triangle D B C$, by the Pythagorean Theorem, we have

$$
B C=\sqrt{B D^{2}-D C^{2}}=\sqrt{3^{2}-1^{2}}=\sqrt{8}=2 \sqrt{2}
$$

Therefore, $A D=2 \sqrt{2} \tan 2 \theta-1$.

But $\tan \theta=\tan (\angle D B C)=\frac{D C}{B C}=\frac{1}{2 \sqrt{2}}$.

This means that $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{1 / \sqrt{2}}{1-(1 / 8)}=\frac{1 / \sqrt{2}}{7 / 8}=\frac{8}{7 \sqrt{2}}$.

Finally, we obtain $A D=2 \sqrt{2} \tan 2 \theta-1=\frac{16 \sqrt{2}}{7 \sqrt{2}}-1=\frac{16}{7}-1=\frac{9}{7}$.

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