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In junior high school mathematics, the steps of solving geometric problems of circular functions and trigonometric functions are detailed and standardized.
1. The idea is easy to understand. Let me reverse the problem solving process first.

To prove EG=AF, just prove that angle EDG= angle ADF.

Because the angle ADB=90 degrees, it is only necessary to prove that the angle EDF=90 degrees.

The following is the main process,

Because AFDE is a circle with four points, ∠ A = ∠ FDE = 90.

Another method is because ∠ A = 90, so EOF is the diameter, so EDF=90.

BAD angle = 45, so EOD angle = 90, so EOF is a bisector, so ADB = 90, ADF=EDG, because ADB = 90. So EG=AF

2。 Connecting EF, OC, OC and FD to H. ADE = FDC is easy to get from the first question.

It is easy to know that OC bisects FD vertically. In the isosceles right triangle EFD, it is known that EF is equal to the diameter, so FD can be obtained, and semi-HD of FD can be obtained, and CD is easy to obtain. In the right triangle CDH, two sides are known, and tan∠FDC can be obtained. Equal to asking.

It is troublesome to work out the result.