Chords in a Circle

Since $\overline{AB}$ bisects $\overline{CD}$ , $CP=PD=6$ . By power of a point, $AP\cdot PB=CP\cdot PD=6\cdot 6=36$ Since $AP+PB=12$ and $AP\cdot PB=36$ , $AP=PB=6$ . Since $PA=PB=PC=PD=6$ , $A$ , $B$ , $C$ , and $D$ lie on a circle centered at $P$ with radius 6. However, since three non-collinear points define a circle, and $A$ , $B$ , $C$ , and $D$ already lie on a circle centered at $O$ , $O$ and $P$ coincide, and $OP=\boxed{0}$ .

Since $AB$ bisects $CD$ , center $O$ must lie in $AB$ . Since, $AB=CD$ , $CD$ bisects $AB$ too. Hence, they intersect at center $O$ , i.e. they both are the diameter of the circle. And hence $O$ and $P$ are the same point, implying $OP$ to be $\fbox{0}$ .

o and p coincide

As it is given that AB=CD+12 cm. So we know that equal chords are equidistant from the center. Therefore the points O(center of the circle) and P the intersection points of the two chord are same.So distance between them is 0.

We see that for the two cords to bisect each other, they must meet at the middle, P, thus OP equals 0.

If 1 chord bisects an other chord,it has to pass through the center of the circle. Now if both are equal, they are diameters of the circle.2 diameters intersect at center i.e, o.So, o and p are the same.