The perpendicular of the intersection point D of AB intersects with the point G, and the bisector of AB is distributed as m and n, as shown in the figure:

∠C = 90∠A = 30∠ABD =∠DBC = 30∠DGB = 90∠ADB = 120 =∠MDE =∠ND(E)？

△DCB and△△ dgb are congruent triangles (db = db ∠ c = ∠ dgb = 90; ∠ABD=∠DBC=30？ )

∠ADG=∠BDG=∠CDB=60

△ADB is an isosceles triangle, and M and N divide AB equally.

Let AD=6x, then DG = 3x and Ag = 3 √ 3x.

So AB=6√3x

AM=MN=NB=2√3x

MG=√3x=NG

DG=3x, MG=√3x at right angles △DGM.

Then DM=2√3x

So ∠ MDG = 30.

Similarly, ∠ NDG = 30.

So ∠ ADM = ∠ MDG = ∠ GDN = ∠ NDB = ∠ BDE = ∠ EDC = 30.

And it can be found that point (e) and point (c) are the same.

So ∠ bed = 120? Still 90?