In the diagram, $\triangle A B C$ is right-angled at $C$. If $B D=2 x, D C=x$, and $\angle A D C=2(\angle A B C)$, then the length of $A B$ is
(A) $2 \sqrt{2} x$
(D) $3 x$
(B) $\sqrt{6} x$
(C) $2 \sqrt{3} x$
(E) $4 x$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-124.jpg?height=276&width=333&top_left_y=1087&top_left_x=1319)
Solution 1

Let $\angle A B C=\theta$. Then $\angle A D C=2 \theta$, and so $\angle A D B=180^{\circ}-2 \theta$ and $\angle B A D=\theta$.

Thus $\triangle A D B$ is isosceles with $B D=D A$, and so $D A=2 x$.

Since $A D$ is twice the length of $D C$ and $\triangle A D C$ is rightangled, then $\triangle A D C$ is a $30-60-90$ triangle, that is, $\angle A D C=60^{\circ}$ and so $\angle A B C=30^{\circ}$.

Therefore, $\triangle A B C$ is also a $30-60-90$ triangle, and so $\angle B A C=60^{\circ}$ (which is opposite side $B C$ of length $3 x$ ).

Thus, $A B=\frac{2}{\sqrt{3}} B C=\frac{2}{\sqrt{3}}(3 x)=2 \sqrt{3} x$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-242.jpg?height=360&width=436&top_left_y=243&top_left_x=1365)

## Solution 2

Let $\angle A B C=\theta$. Then $\angle A D C=2 \theta$, and so $\angle A D B=180^{\circ}-2 \theta$ and $\angle B A D=\theta$.

Thus $\triangle A D B$ is isosceles with $B D=D A$, and so $D A=2 x$.

Since $\triangle A D C$ and $\triangle A B C$ are both right-angled, then

$$
\begin{aligned}
A B^{2} & =B C^{2}+A C^{2} \\
& =B C^{2}+\left(A D^{2}-D C^{2}\right) \\
& =(3 x)^{2}+\left((2 x)^{2}-x^{2}\right) \\
& =9 x^{2}+3 x^{2} \\
& =12 x^{2}
\end{aligned}
$$

and so $A B=\sqrt{12} x=2 \sqrt{3} x$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-242.jpg?height=357&width=441&top_left_y=800&top_left_x=1357)