✔ 最佳答案
Because AD=DB, AE=CE, by mid-pt theorem,
(1) (a) BC=2DE, but FG= 1/3BC=2/3DE
so 2DE=3FG
(b) By mid-pt theorem, DE//BC
Angle FHG= Angle DHE (common)
Angle HFG= Angle HDE (alt Angle DE//BC)
Angle HGF= Angle HED (alt Angle DE//BC)
So Triangle HFG is similar to triangle HDE (AAA)
HG/HE = FG/DE=2/3 , so HG:GE=2:1 (corr side of similar triangle)
Similarly HF:FD=2:1 (intercept theorem)
(c) Angle CGE= Angle HGB (vert opp angle)
EG:HG= 1:2 by (b)
Because CG:CB=1:3, so CG:BG=1:2
so TriangleCEG~TriangleBHG (ratio of 2 sides, inc angle)
(d) Angle ECG = Angle GBH (corr angle of similar triangle)
so EC//BH ( alt angle equal)
EG/HG=CE/BH=1/2 (corr side of similar triangle)
so BH=2CE, but AC=2CE , so BH=AC
so ABHC is a //gram (2 sides equal and //)
(2) Let the side of the square base be x cm
(√2x^2) =12√2 , x=12
Height of the 4 triangles= (√6^2+8^2)=10cm
Total surface area of pyramid= 1/2(12)(10)(4)+12^2= 384cm^2