Only one person (I am currently in the third grade, taking the senior high school entrance examination) practices: (the method of translation (1))

Direct (2): Let NH be perpendicular to BC, connect NC, let NK be parallel to BC, let EK be perpendicular to NK, let BC be in Q, let NK pass AB to G and connect QN.

NH is the midpoint and NC = BN = NE. Because CQ = EQ (45 degrees), NQ is the center line of CE. The triangle HNC that is easy to prove is equal to the triangle NEK, and the angle BNE of the triangle BNG that is easy to prove is 90 degrees.

Let NK pass through CF in I and then CE+IF = radical 2cd/2 and then CE = radical 2/2 (CD-2ga).

Because of the neutral line, ah = 1/2am, MB = CD-2ga.

So ce/ab = root number 2/2