(Prove) The angle subtended by an arc at the center is double the angle subtended by it at anypoint on the remaining part of the circle.

Sure, here is the proof:

Step 1: Let's consider a circle with center O. Let AB be an arc and let's consider two points C and D on the circle such that they subtend ∠COB at the center and ∠DAB at the remaining part of the circle.

Step 2: Draw the radii OC and OD. Also, draw OB.

Step 3: In triangle OCB, ∠COB is an angle subtended by the arc CB at the center.

Step 4: In triangle ADB, ∠DAB is an angle subtended by the same arc CB at the remaining part of the circle.

Step 5: Now, ∠COB = 2 * ∠CAB. This is because the angle subtended by an arc at the center is double the angle subtended by it at any point on the alternate segment of the circle.

Step 6: Therefore, ∠COB = 2 * ∠DAB.

Hence, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.