As shown in the figure, in the isosceles right triangle ABC, ∠ ACB = 90 °, point D is the midpoint of BC, and the isosceles right triangle DCE is made with CD as the side, where the angle DCE = 90 ° CD = CE, link AE, the quantitative relationship between AE and BD, explain the reason

As shown in the figure, in the isosceles right triangle ABC, ∠ ACB = 90 °, point D is the midpoint of BC, and the isosceles right triangle DCE is made with CD as the side, where the angle DCE = 90 ° CD = CE, link AE, the quantitative relationship between AE and BD, explain the reason

1. Prove: in equilateral triangle ABC, AC = CB, angle ACB = B, and CD = BF, so triangle ACD is equal to triangle CBF2. When D is at the midpoint of BC, prove: at this time, point F is also at the midpoint of AB, so angle BCF = angle CAD = 30 degrees, angle ADB = 90 degrees. In equilateral triangle ADF, ad = De, angle ade = 60 degrees, so angle BDE = 30

It is known that: as shown in the figure, △ ABC, AC = BC, ⊙ o with BC as the diameter intersects AB at point D, de ⊥ AC is made at point e through point D, and the extension of BC intersects at point F. it is proved that: (1) ad = BD; (2) DF is the tangent of ⊙ o

It is proved that: (1) connecting CD, ∵ BC is the diameter of ⊙ o, ∵ CD ⊥ ab. ∵ AC = BC, ∵ ad = BD. (2) connecting OD; ∵ ad = BD, OB = OC, ∵ od is the median line of ⊙ BCA, ∵ OD ∥ AC. ∵ de ⊥ AC, ∵ DF ⊥ OD. ∵ od is the radius and ⊙ DF is the tangent line of ⊙ o