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Kneel for the solution to the math finale of Shaoxing senior high school entrance examination today (the second small question)
There is a good method, because ∠ CDE = ∠ CQE = 90, so C, E, Q and D are all on the circle with the diameter of CE, so ∠DEC=∠DQC= semi-arc CD.

Then you can ask. That is, ∠ dqc = 45, so △BCQ is isosceles Rt△, so CQ=BC=3, so q (4 4,0) can be solved analytically.

Secondly, the △BCQ proved by the above method is Rt△ band 30, but it is divided into 1 groups on the left and right sides of the symmetry axis, and the 30 angle of each group is rotated to C and E, that is, * * *. There are four cases.

Papers and answers will be sent out immediately, with detailed procedures (but unlike mine, I still feel simple, (* _ *) hee hee ...)