In the diagram, $A B$ is perpendicular to $B C$, and $C D$ is perpendicular to $A D$. Also, $A C=625$ and $A D=600$. If $\angle B A C=2 \angle D A C$, what is the length of $B C$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_5629b903ac0448ea13d7g-2.jpg?height=326&width=419&top_left_y=1908&top_left_x=1357)
By the Pythagorean Theorem,

$$
C D^{2}=A C^{2}-A D^{2}=625^{2}-600^{2}=390625-360000=30625=175^{2}
$$

Since $C D>0$, we get $C D=175$.

Let $\angle D A C=\theta$. Thus, $\angle B A C=2 \angle D A C=2 \theta$.

Looking in $\triangle A D C$, we see that $\sin \theta=\frac{C D}{A C}=\frac{175}{625}=\frac{7}{25}$ and $\cos \theta=\frac{A D}{A C}=\frac{600}{625}=\frac{24}{25}$. Thus,

$$
\sin 2 \theta=2 \sin \theta \cos \theta=2 \cdot \frac{7}{25} \cdot \frac{24}{25}=\frac{336}{625}
$$

Since $\sin (\angle B A C)=\frac{B C}{A C}$, then

$$
B C=A C \sin (\angle B A C)=625 \sin 2 \theta=625 \cdot \frac{336}{625}
$$

and so $B C=336$.