Circular reasoning

We note that $AF=BO = 5$ , $OF=OC=r$ and $\angle OAF = \angle CBO = 90^\circ$ , therefore $\triangle AOF$ and $\triangle CBO$ are congruent. So $AO=CB$ or $h=\boxed{12}$ .

If we consider the lower triangle, we can form a 5-12-13 right triangle. Thus, the radius is 13. Now in the upper triangle, we can form another 5-12-13 triangle. Thus, h = 12.

(r-5)(r+5)=12×12 ............ r^2 - 25=144 ................... r =13 .............. From pythagorean theorem h^2=r^2-5^ Then h=12

The triangle DO and right angle = triangle OE and right angle. Therefore, h = 24/2 = 12.

The triangles E-O-1/2EF and D-O-1/2DC are equal as they both consist of sides 5 and a radius, therefore we have h given as 1/2 DC or 12.

5^2+12^2=5^2+h^2

I) OD = OC = OE = OF as radii of same circle.
II) According to Pythagoras theorem OD = 13, as OD^2 = (5)^2 + (24/2)^2
III) In triangle OEF, OE = OF = 13 and half of base = 5. Let 'A' be the mid-point of line EF i.e. base of triangle OEF. By applying Pythagoras theorem on triangle OAF will get OA i.e. h = 12.

Let us consider the mid point of DC be A and the mid point of EF be K Considering triangle OAC, we have (OC)^2 = (OA)^2 + (AC)^2 and we get OC=13. Since O is the centre of the circle, OC and OF are the radius and hence they are equal. Thus in triangle OKF, (OK)^2 = (OF)^2 + (KF)^2 And thus h = 12

Notices that there are 4 right triangles and they satisfy 5^2 +12^2=13^2. So, h=12.