Angel AHC= angel DBC

Re-proof that triangle CFH is equal to triangle CGB

Get CF=CG and angle FCG=60 degrees.

equilateral triangle

So GFC angle DCA angle =60 degrees.

So FG is parallel to AB

2. extend Ba and Ce to point p and intersect with point d to make DH perpendicular to BC to point H.

Using angular bisector BD

Prove that all bad triangles are equal to BHD of triangles.

Triangle BEP is equal to triangle BEC.

Used AB=BH=AC, PC=2EC.

Angle DBH=22.5 degrees = angle alpha can be easily obtained by using an angle of 45 degrees.

So the triangle BHD is equal to the triangle ACP(ASA).

BD=PC=2EC

3( 1). The intersection D is DM, parallel to AC, and AB is at point M.

An equilateral triangle MBD can be obtained.

AM=DC, angle AMD= angle DCE= 120 degrees.

Because angle ADC= angle B+ angle BAD

Angle MAD= angle EDC

Triangle AMD is equal to triangle DCE.

AD=DE

(2) Adding straight lines in the same way can prove that all triangles AMD are equal to triangles DCE.

Only point m is on the AB extension line.