AB=x 1-x2=√5

(x 1-x2)^2=(x 1+x2)^2-4x 1x2=m^2+4(-m+2)=(√5)^2=5

m^2-4m+3=0

(m- 1)(m-3)=0

M= 1 or m=3

And because a and b are on both sides of the origin.

Therefore, x 1 * x2 = m-2.

m & lt2

So, m= 1

Solution: the second problem

The intersection of CH parallel AF c, g crosses DE and h crosses AD.

Because e and f are the midpoint, ABCD is the square, and h is the midpoint of AD,

So ∠ DAF = ∠ AFB = ∠ AED, DE⊥AF, HC⊥DE.

△DAP∽△DEA∽△DHG

And because 2ae = ad, so

G is the midpoint of PD,

CH is the middle vertical line of DP,

△CDP is isosceles△, and the vertex is c,

So DC = PC = 4,

ABCD area = 4 * 4 = 16