Triangle ABC, triangle ace and triangle BCF are equilateral triangles with AB, AC and BC sides of triangle ABC respectively. Quadrilateral ADFE is a parallelogram Give reasons

Triangle ABC, triangle ace and triangle BCF are equilateral triangles with AB, AC and BC sides of triangle ABC respectively. Quadrilateral ADFE is a parallelogram Give reasons

Is the title wrong? What about point d?
It should not be a parallelogram. You can see it when you draw a little more standard

In addition to the triangle ABC, we take AB and AC sides as equilateral triangle abd and equilateral △ ace respectively to prove DC = be

Because of the equilateral triangle abd and equilateral △ ace, the angles EAC = DAB = 60 degrees, ad = AB, AC = AE
So angle EAB = DAC, so triangle DAC is all equal to BAE, (SAS), so DC = be

In △ ABC, ∠ BAC = 90 °, ab = AC, point D is the middle point of BC side, AC is the hypotenuse to make a right triangle ace, ∠ AEC = 90 ° to connect De
1. If △ AEC is outside △ ABC, we prove that AE + CE = root 2DE
2. If △ AEC is in the interior of △ ABC, it is to judge that the quantitative relationship of AE, CE and De is [] and prove that

According to the meaning of the title, take the midpoint o of AC as the center of the circle and make a circle with the diameter of AC
Obviously, a, D and C are all on the circle, and there are some problems
If f, the midpoint of DC, connects fo and extends the intersection with the circle at e, it is obvious that de = EC
It contradicts the conclusion of question 1,

It is known that the bisector AE of ad ⊥ DC, BC ⊥ CD ∠ DAB and bisector be of ABC intersect at point E, and point E is on DC. Prove de = CE
！！！！！！！！！！！！！！！！！！！！！！！！！！！！！ What about AB = AD + BC?

Passing through point E as EM ⊥ AB in M
∵ AE bisection ∠ DAB, ∵ EM = de
∵ EB bisection ∠ ABC, ∵ EM = CE
∴DE=CE