So the answer is: △ ADC can be obtained by rotating △ABC around point O by180.
(2) As shown in Figure 2-2- 1, connect BB' and EF vertically divide BC.
∴BB'=B'C, which can be obtained by folding.
∴B'C=BC，
∴△BB'C is an equilateral triangle.
∴∠B'CB=60，
∴∠B'CG=30，
∵∠GB′C = 90，
∴∠b'gc=60；
(3) As shown in Figure 3- 1- 1, take the midpoints P, Q and R of CE, EG and GI respectively and connect DP, FQ, HR, AD, AF and AH.
∵BA=BC, according to the nature of translation transformation,
∴△CDE, △EFG and △GHI are all isosceles triangles.
∴dp⊥ce,fq⊥eg,hr⊥gi,gr=eq=cp=0.5a,dp=fq=hr.
AC = a，
∴AI=4a.
AH = AI，
∴AH=4a,AR=3.5a.
∴AH2= 16a2.
In Rt△AHR, AH2=HR2+AR2,
16a2=HR2+494a2，
HR2= 154a2，
∴DP2=FQ2=HR2= 154a2，
In Rt△ADP and Rt△AFQ, from Pythagorean theorem, we get
AD2=AP2+DP2=6a2，AF2=AQ2+FQ2= 10a2，
∴AH2=AD2+AF2，
The new triangle is a right triangle,
The three sides of the new triangle are 4a, 6a and 10a. ..
Its area is:126a×10a =15a2.
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