Answer:
Explanation:
To show that point O is the center of the circle that passes through points D, E, and F, we can use the property of perpendicular bisectors. The perpendicular bisectors of two chords of a circle bisect the chord and pass through the center of the circle. Since the perpendicular bisectors of DE and EF intersect at point O, it follows that O must be the center of the circle that passes through points D, E, and F.
This property can also be proven using the Pythagorean theorem. Let the radius of the circle be r and let the midpoints of DE and EF be M1 and M2, respectively. Then, OM1 = OM2 = r, and the distance between M1 and M2 is equal to the distance between D and F. By the Pythagorean theorem, the sum of the squares of the sides of a right triangle is equal to the square of the hypotenuse, so we have:
OM1^2 + (M1M2)^2 = r^2 + r^2 = 2r^2
Since OM1^2 + (M1M2)^2 is equal to 2r^2, it follows that the distance between M1 and M2 is equal to the diameter of the circle. This means that M1 and M2 are the midpoints of the diameter of the circle, so they must both lie on the circumference of the circle. This, in turn, means that O is the center of the circle that passes through points D, E, and F.