We are given a figure containing a triangle $BCD$ with $\angle B = 25^{\circ}$ and $\angle D = 90^{\circ}$. We are asked to complete the proportion $\frac{BD}{BC} = \frac{?}{CA}$ based on the figure.

1. Problem Description

We are given a figure containing a triangle $BCD$ with $\angle B = 25^{\circ}$ and $\angle D = 90^{\circ}$. We are asked to complete the proportion $\frac{BD}{BC} = \frac{?}{CA}$ based on the figure.

2. Solution Steps

First, consider the given triangle $BCD$. Since the sum of angles in a triangle is $180^{\circ}$, we have

$$\angle BCD = 180^{\circ} - \angle B - \angle D = 180^{\circ} - 25^{\circ} - 90^{\circ} = 65^{\circ}.$$

The proportion $\frac{BD}{BC} = \frac{?}{CA}$ involves sides $BD$ and $BC$ of the triangle $BCD$, as well as $CA$. We are looking for a triangle that involves the side $CA$. Since the line $BC$ is extended, let us consider the triangle $ACD$.

We know that $\angle DCA = 90^{\circ}$, thus the triangle $ACD$ is a right triangle.

Consider the proportion

$$\frac{BD}{BC} = \frac{AD}{AC}.$$

Since $\sin(\angle B) = \frac{opposite}{hypotenuse}$, we have $\sin(25^{\circ}) = \frac{CD}{BC}$ in the triangle $BCD$.

Also, $\cos(\angle B) = \frac{adjacent}{hypotenuse}$, which gives $\cos(25^{\circ}) = \frac{BD}{BC}$ in the triangle $BCD$.

In triangle $ACD$, $\tan(A) = \frac{CD}{AC}$.

In triangle $ABC$, $m\angle BAC = 90 - 25 = 65$.

Then $\angle BCA$ is a right angle and $ACD$ is a right angle. So both $\angle DCA$ is $90^{\circ}$ and $\angle BCA$ is $90^{\circ}$.

Let $\angle ACD = x$. In $\triangle ACD$, $\angle DAC + \angle ACD = 90$, so $\angle DAC + x = 90$, so $\angle DAC = 90 - x$.

In triangle $BCD$, we have that $\cos(25) = \frac{BD}{BC}$. We seek the equality

$$\frac{BD}{BC} = \frac{CD}{AC}$$

So, the proportion becomes

$$\frac{BD}{BC} = \frac{AD}{AB}$$

The proportion $\frac{BD}{BC} = \frac{CD}{CA}$ is incorrect.

In the triangle $BCD$, $\sin(B) = \frac{CD}{BC}$, thus $CD = BC\sin(B)$.

$\cos(B) = \frac{BD}{BC}$, thus $BD = BC\cos(B)$.

Substituting $CD$ and $BD$ in the given proportion results in $\frac{BC \cos(B)}{BC} = \frac{CD}{CA}$.

We have $\cos(B) = \frac{CD}{AC}$. This doesn't work.

Consider the triangle $BDA$. Since $D$ is $90$, $\triangle ABD$ is not necessarily similar to anything.

However, note that since we have a right angle at $C$, the angle formed, $\angle BCD$ is also $90$, so we need to create similar triangles. So

$$\frac{BD}{BC} = \cos(25)$$

$$\frac{CD}{BD} = \tan(25)$$

$$\frac{BD}{CD} = \cot(25)$$

Consider triangles $BDC$ and $ABC$. If they were similar triangles, then $\angle B$ would have to be congruent, so $\frac{BD}{BC} = \frac{BC}{BA}$.

$\angle DAC = 65$

$\angle ACD = \frac{\angle A}{90^\circ}$. So what should be put here?

In $\triangle ABC$, $\angle A = 180-90-25 = 65$. Also $\sin(A)=\frac{BC}{AB}$. and $\frac{BD}{BC} = \frac{AC}{AB}$. So we should instead consider $\triangle ABC$.

3. Final Answer

AC

AB