Lines, angles, and triangles SAT question

Choice D is correct. It's given that in triangle $ABC$, the measure of angle $B$ is $90^{\circ }$ and  $BD$ is an altitude of the triangle. Therefore, the measure of angle $BDC$ is $90^{\circ }$. It follows that angle $B$ is congruent to angle $D$ and angle $C$ is congruent to angle $C$. By the angle-angle similarity postulate, triangle $ABC$ is similar to triangle $BDC$. Since triangles $ABC$ and $BDC$ are similar, it follows that $\frac{AC}{AB}=\frac{BC}{BD}$. It's also given that the length of $\overline{\mathrm{AB}}$ is $15$ and the length of $\overline{\mathrm{AC}}$ is $23$ greater than the length of $\overline{\mathrm{AB}}$. Therefore, the length of $\overline{\mathrm{AC}}$ is $15+23$, or $38$. Substituting $15$ for $AB$ and $38$ for $AC$ in the equation $\frac{AC}{AB}=\frac{BC}{BD}$ yields $\frac{38}{15}=\frac{BC}{BD}$. Therefore, the value of $\frac{BC}{BD}$ is $\frac{38}{15}$.

Choice A is incorrect. This is the value of $\frac{BD}{BC}$.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.