Two. In a regular pentagon, Mn = NP = PQ = QD = DM = M, AD = BC = C'd = B, AB = CD = A.

∠DMN =∠MNP =∠NPQ =∠PQD =∠QDM = 108

① From the folding process, DE=BE, ∠ ADE = ∠ CDF = ∠ QDM-∠ ADC =108-90 =18.

There is AE/AD=tan∠ADE in Rt△ADE, that is, AE = ad * tan ∠ ade = b * tan 18.

Answer? -B? =(AE+EB)？ -Advertising?

=(AE+ED)？ -Advertising?

=AE？ +2AE*ED+ED？ -Advertising?

=2AE？ +2AE*ED

=2AE(AE+ED)

=2AE(AE+EB)

=2*b tan 18 *a

=2ab tan 18

② If DN is connected, then ∠ NDG = ∠ MDG-∠ MDN = ∠ QDM/2-(180-∠ DMN)/2 =18.

GN = NP/2 = m/2，DG = BG = BD/2 = √(a？ +b？ )/2

In Rt△DNG, tan ∠ ndg = ng/DG = (NP/2)/(BD/2) = NP/BD.

That is, m = NP = BD tan∠NDG = √(a? +b？ )tan 18

3 even billions.

AE and A'E' are symmetrical about MN, AE intersects with A'E' at point B, so B is on the symmetry axis MN, that is, the BMN*** line.

In Rt△ABM, ∠ ABM = 90-∠ AMB = 90-(180-108) =18.

tan∠ABM = AM/AB = (AD-DM)/AB

b = AD

= DM + AB tan∠ABM

= m + a tan 18

④ If NH⊥MQ is in H, then ∠ MNH = ∠ MNP-∠ PNH =108-90 =18.

In Rt△MNH, tan∠MNH = MH/NH.

A'H = NG = NP/2 = m/2

There is MH = mnsin ∠ mnh = msin 18.

b = AD

= DM + AM

= Deutsche Mark+AM

= DM + A'H + MH

= m + m/2 + m sin 18

= (3/2)m + m sin 18

& lt(3/2)m + m tan 18

Therefore, ① ② ③ holds, ④ does not.