∫ two congruent trigonometric rulers ADE and ABC, with angles of 30 and 60 respectively,

∴DE⊥EC,BC⊥EC,DE∥BC

M is the midpoint, and the vertical line MF⊥EC is taken from M.

∴DE∥MF∥BC and EF=FC.

∴△EFM and△△△ CFM are congruent,

Get EM=MC

So △EMC is an isosceles triangle.

2. extend AE to g, so that EG=AE, even DG.

Because AE = eg, DE = EC and angle DEG = AEC, triangle DEG is congruent with AEC.

Because df = AC = DG and angle g = angle DFE,

Because of parallelism, angle BAF= = angle DFE= = angle G = angle EAC, AE divides BAC equally.