∵ quadrilateral ABCD and quadrilateral CEFG are both squares.
∴BC=DC,CG=CE,∠BCG=∠DCE=90
∴△BCG∽△DCE
Therefore, BG = ce, ∠ bgc = ∠ dec
And < bgc+< cbg = 90.
∴∠DEC+∠CBG=90
The straight line of BG and DE is cut by the straight line of BC, and the inner angle of the same side is the complementary angle, then the straight line BG⊥DE.
(2) it still holds.
Proof: As shown in Figure 2, ∠ BCD = ∠ GCE = 90 in square ABCD and CEFG.
∠∠BCD+∠DCG =∠GCE+∠DCG, that is ∠ BCG = ∠ DCE = 90.
BC=CD，CG=CE
∴△BCG∽△DCE
∴BG=CE,∠CBG=∠CDE
And ≈CBG+∠ bhc = 90.
∴∠CDE+∠BHC=90
Then the straight line between BG and DE is cut by DC, and the inner angle on the same side is complementary angle, and there is a straight line BG⊥DE.
(2) As shown in Figure 5, in rectangles ABCE and CEFG, ∠ BCD = ∠ GCE = 90.
∠∠BCD+∠DCG =∠GCE+∠DCG，
Yes, BCG = DCE = 90.
AB = a，BC=b，CE=ka，CG=kb
BC/CD=b/a，CG/CE=kb/ka=b/a
∴△BCG∽△DCE (two triangles with equal angles and proportional sides are similar)
Yes ∠CBG=∠CDE
≈CBG+≈ bhc =90
∴∠CDE+∠BHC=90
Then the straight line between BG and DE is cut by DC, and the inner angle on the same side is complementary angle, and there is a straight line BG⊥DE.
Similarly, in right-angle ABCE, A and B are not equal?
The ratio of b to a is not 1, and BG is not equal to DE.
Therefore, the conclusion in (1) is still valid only if BG and DE are perpendicular to each other.