請數學高手!(四邊形問題ENG)

Let 2a be the length of length of each side of thesquares ABCD and EFGH.

P is the mid-point of CD and EF, and Q is the mid-point of AD and EH.
AQ = QD = EP = PF = EQ = QH = CP = PD = a

EQ = QD = DP = PE = a
∠QDP = ∠PEQ = 90° (int. ∠ of sq.)
Hence, EQDP is a square.

For sq. ABCD : BA // CD and BC // AD (opp. sides of sq.)
For sq. EFGH : EH // FG and EF // HG (opp. sides of sq.)
For sq. EQPD : EH // CD and EF // AD (opp. sides of sq.)
Hence, BA // EH // CD // FG and BC // EF // AD // HG

Produce AD to meet FG at M, and produce CD to meet HG and N.
QD // HN (proven)
QH // DN (proven)
∠QDN = 180° - ∠PDQ = 180° - 90° = 90° (∠s on a st. line)
Hence, QDNH is a sq., and DN = HN = a
Similarly, PDMF is a sq., and DM = FM = a
Similarly, DNGM is a sq., and NG = MG = a

AM = CN = 2a + a (proven)
MG = FM = a (proven)
∠CNG = ∠AMF = 90° (int. ∠ of sq.)
∆CNG ≡ ∆AMF (SAS)

Let X be the point of intersection of AF and CD.
Consider ∆ADX and ∆COX.
∠XCO = ∠XAD (corr. ∠s, ∆CNG ≡ ∆AMF)
∠AXD = ∠CXO (vert. opp. ∠s)
Since the two ∆s have 2 ∠s equal, the 3rd ∠ must be equal.
Hence, ∠COX = ∠ADX = 90°
Hence, AF⊥CG

Constructions :
(1) BA produce to meet GH produce at M.
(2) BC produce to meet GF produce at N.
(3) AD produce to meet GF at X.
(4) Join MX.
Results :
(1) BMGN is a square.
(2) Triangle AFX congruent triangle MXG. ( SAS). Therefore AF//MX.
(3) Triangle CNG congruent triangle MXG. (SAS)
Therefore angle CGN = angle XMG
Let MX and CG intersects at K
Angle CGN + angle MGK = 90 degree
that means angle XMG + angle MGK = 90 degree
so angle MKG = 90 degree ( angle sum of triangle MKG)
therefore angle AOG = 90 degree (AF//MX)
so AF is perpendicular to CG.