The question is as follows:
The length of segment AB is 20 cm. Find the distance from C to AB, given that C is a point on the circle that has AB as a diameter, and that
(a) $AC$ = $CB$; (b) $AC$ = 10 cm; (c) $AC$ = 12 cm
According the Thales' Theorem, if $A$, $B$, and $C$ are distinct points on a circle where the line $AB$ is the diameter of the circle, then $\angle ABC$ is a right angle.
For (a) I was thinking of right triangle $ABC$, where $\angle ACB$ is a right angle and the diameter $AB$ is the hypotenuse of the right angle. Since $AC = BC$, it shows that it is a 45-45-90 triangle because of its side lengths $x- x- x\sqrt{2}$, and I got $x = 10\sqrt{2}$cm. However, I was thinking that $10\sqrt{2}$cm is only the measure from $C$ to the points $A$ and $B$, so to actually get from $C$ to $AB$, I thought of creating another right triangle where the right angle would be on the intersection from the line drawn from $C$ to the diameter and the hypotenuse would be $10\sqrt{2}$cm and the measure of a leg would be $10$cm (half of the diameter) and I would solve for the other leg, which I got to be $10$cm as well. Is this approach correct?
Also, for (b), I was only able to get to finding the measure of $BC$ to be $10\sqrt{3}$cm, but I don't know if I should do additional steps to find the distance from $C$ to the diameter. I thought of doing $sin(60) = \frac{d}{10}$, where $d$ is the measure of the distance from $C$ to $AB$, and I got $d \approx 8.66$cm.
I believe part (a) is simpler than you think. It is given that $AB$ is a diameter. If $AC=AB$, then $AC$ must be a diameter as well. For any point on a circle, there is only one other point on the circle such that the line segment between the two points is a diameter. Thus, for part (a), the distance is 0.
For part (b), you are on the right track - it is indeed a right triangle. In fact, it is a special right triangle - a 30-60-90. So you just need to find the altitude of the triangle. There is an easy way to do this. The area is half the product of an altitude and a base. So if the altitude from the hypotenuse to the base is $\alpha$, we have $$\frac{\alpha \cdot AB}{2} = \frac{AC \cdot BC}{2}$$ and therefore $$\alpha = AC \cdot BC / AB.$$ And of course since it is a 30-60-90 triangle, $$BC = 10\sqrt{3}.$$
Part (c) is essentially the same, except with a 12-16-20 (which is a 3-4-5 triangle) right triangle. So in this case we have $AC = 12, BC = 16, AB = 20$, and the same formula from above applies for the closest distance (altitude) $\alpha$.
Hope this helps.