The trouble on the second floor is terrible, equations, auxiliary lines, Pythagorean theorem!

I'll tell you a very simple method, which can be obtained in D after two congruences.

Immortal

DF rotates the triangle DAE counterclockwise by 90 degrees, so that AD and DC coincide (the figure should be similar), and point E is point H. If you draw in the opposite direction, rotate it clockwise.

In this way, all triangles DCH are equal to triangles ADE, so DE=DH and AE=CH, so HF=CF+CH=CF+AE=EF and DF=DF.

So SSS can prove that triangle DEF is equal to triangle DFH,

Because the corresponding heights of congruent triangles are equal (I don't know, you can also use the method of area and bottom edge and area)

So DG=DC

So DG=DC=DA.