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Problems in mathematical circles
1. Extend EA, intersect M, and connect BM. Let BF intersect with n and connect an.

AB is the diameter, so the angles AMB and ANB are right angles.

ANFE and MBFE are two rectangles.

Finally, EC*ED=EA*EM=FN*FB=FD*FC.

Get EC=DF.

2

ACB right triangle, AC*BC is twice the area of the triangle.

Passing c is high, and H * 2oc = AC * bc = oc 2.

Get h=OC/2.

Observe that the hypotenuse of the triangle CHO (drawn by yourself) is OC, which is twice as high as H, so the angle COH=30 degrees.

So CBA=30/2= 15.

Cab =75