From what I've understood, Since BC is a diameter, we have $ \angle CDB=90 $. From the fact that $ \angle CAB = 46 $ we have that $\angle DBC = 44 $ and $ \angle DCB = 46 $. By Adding Angles, we get that since $\angle ACB $ is right, $\angle FCD $ is 44. Since F is a common tangent to the given circle, FD=FC, or in other words, $ \angle DCF = \angle CDF $ Thus, $ \angle CFD = 180 - 88 = 92 $ degrees ^ The wording is a bit confusing, so please correct me if I misunderstood it!

Leicich wrote: 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 $\angle{CFD}$. Assuming that the tangent is from $D$, we have $AC$ to be tangent too $\implies FC$ is also tangent to this circle. By alternate segement theorem, we get $\angle FDC=\angle CBD = 44^\circ$. Also, as $FC=FD$, we get $\angle CFD=92^\circ$. $\blacksquare$

Mr.Bash wrote: Leicich wrote: 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 $\angle{CFD}$. Assuming that the tangent is from $D$, we have $AC$ to be tangent too $\implies FC$ is also tangent to this circle. By alternate segement theorem, we get $\angle FDC=\angle CBD = 44^\circ$. Also, as $FC=FD$, we get $\angle CFD=92^\circ$. $\blacksquare$ Lol this is way simpler!