Final answer:
D is the center of the circle as it lies on the perpendicular bisector of AB, making it equidistant from the vertices of triangles CAD and CBD, which define the circle.
Step-by-step explanation:
To prove that point D is the center of the circle given triangles CAD and CBD, we rely on the fact that AD=DB. If AD=DB, by definition, D lies on the perpendicular bisector of segment AB. This implies that D is equidistant from A and B. Since this is true for a particular segment of a triangle, and since D lies on the circumcircle of triangle CAD and CBD (by definition of a triangle's circumcircle), it would hold true for all points on the triangle. Therefore, triangles GFC and AHD would also be congruent to the shaded triangles given their side lengths and angles, also confirming that D is equidistant from A, B, and the other vertices of the triangles. Consequently, D is the center of the circle because it is equidistant from all the vertices of the given triangles, which are points on the circle.