Therefore, ∠EFG= 180-75-30=75

Therefore, ∠EFG =∠ epidermal growth factor

It is also easy to get that △EFG is similar to △EHC, and both are isosceles triangles, so FH=GC is obtained.

The quadrilateral FGCH is an isosceles trapezoid.

2)S△EQC = 1/2 * QC * EC = 1/2 * 1 *√3 =√3/2

s△EHC = 1/2 * EC * EH * sin 30 = 1/2 *√3 *√3 * 1/2 = 1/4

S△HQC = S△EQC-S△EHC =√3/2- 1/4 =(2√3- 1)/4

3)QE=2, EH=EC=√3, so QH=2-√3.

Because DG/DC = tan 15, DG = DC * tan15 = tan15.

= tan(45-30)=(tan 45-tan 30)/( 1+tan 45 * tan 30)

=( 1-√3/3)/( 1+√3/3)=2-√3

So DG=QH?

It is also proved that QE = 2 and Eh = EC = √ 3, so QH=2-√3.

The crossing point c is the high CK on QE, and CK = 2 * s △ EQC/QE = (2 * √ 3/2)/2 = √ 3/2.

Because ∠HCQ=90-75= 15 and ∠QCK=30, BC is the bisector of ∠QCK.

△QKC, HC is the angular bisector, which is obtained according to the theorem of angular bisector.

QC/QH = CK/ Hong Kong

1/(2-√3)=√3/2/HK，HK=√3-3/2

Easy to get △HKC is similar to △GDC. CK/CD = Hong Kong/Guangdong

(√3/2)/ 1=(√3-3/2)/GD

GD=2-√3

So DG=QH?

If you can't find the theorem of angular bisector, you can tell me how to prove it.