In a triangle $ABC$ ($\angle{BCA} = 90^{\circ}$), let $D$ be the intersection of $AB$ with a circumference with diameter $BC$. Let $F$ be the intersection of $AC$ with a line tangent to the circumference. If $\angle{CAB} = 46^{\circ}$, find the measure of $\angle{CFD}$.
1. Given that $\triangle ABC$ is a right triangle with $\angle BCA = 90^\circ$.
2. Let $D$ be the intersection of $AB$ with a circumference with diameter $BC$.
3. Let $F$ be the intersection of $AC$ with a line tangent to the circumference.
4. Given $\angle CAB = 46^\circ$, we need to find the measure of $\angle CFD$.

First, we need to understand the properties of the circle with diameter $BC$:
- Since $BC$ is the diameter, any angle subtended by $BC$ on the circumference is a right angle (Thales' theorem).

Next, we use the Alternate Segment Theorem:
- The Alternate Segment Theorem states that the angle between the tangent and chord through the point of contact is equal to the angle in the alternate segment.

Since $F$ is the intersection of $AC$ with a line tangent to the circumference, and assuming the tangent is from $D$:
- $AC$ is tangent to the circle at $F$, implying $FC$ is also tangent to this circle.

By the Alternate Segment Theorem:
- $\angle FDC = \angle CBD$.

Since $\angle CAB = 46^\circ$ and $\angle BCA = 90^\circ$, we can find $\angle ABC$:
\[
\angle ABC = 90^\circ - \angle CAB = 90^\circ - 46^\circ = 44^\circ.
\]

Thus, $\angle CBD = 44^\circ$:
\[
\angle FDC = \angle CBD = 44^\circ.
\]

Since $FC$ is tangent to the circle at $F$, and $FD$ is also tangent to the circle at $D$, we have:
- $FC = FD$ (tangents from a common external point are equal).

Therefore, $\triangle CFD$ is isosceles with $FC = FD$:
\[
\angle CFD = 180^\circ - 2 \times \angle FDC = 180^\circ - 2 \times 44^\circ = 180^\circ - 88^\circ = 92^\circ.
\]

The final answer is $\boxed{92^\circ}$.