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Geometry and Trigonometry Difficulty: Hard

In triangle $A B C$ , the measure of angle $B$ is $90 degree$ and $B D$ is an altitude of the triangle. The length of $A B$ is $15$ and the length of $A C$ is $23$ greater than the length of $A B$ . What is the value of $\frac{B C}{B D}$ ?

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Explanation

Choice D is correct. It's given that in triangle $A B C$ , the measure of angle $B$ is $90 degree$ and $B D$ is an altitude of the triangle. Therefore, the measure of angle $B D C$ is $90 degree$ . 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 $A B C$ is similar to triangle $B D C$ . Since triangles $A B C$ and $B D C$ are similar, it follows that $\frac{A C}{A B} = \frac{B C}{B D}$ . It's also given that the length of $\bar{A B}$ is $15$ and the length of $\bar{A C}$ is $23$ greater than the length of $\bar{A B}$ . Therefore, the length of $\bar{A C}$ is $15 plus 23$ , or $38$ . Substituting $15$ for $A B$ and $38$ for $A C$ in the equation $\frac{A C}{A B} = \frac{B C}{B D}$ yields $\frac{38}{15} = \frac{B C}{B D}$ . Therefore, the value of $\frac{B C}{B D}$ is $StartFraction 38 Over 15 EndFraction$ .

Choice A is incorrect. This is the value of $\frac{B D}{B C}$ .

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

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