D﹙m？ m？ +m﹢？ ﹚

When y=0, x = m (m+ 1)

X =- 1 or x=2m+ 1.

Therefore, a is a constant (﹙ 1, 0 ﹚, b ﹙ 2m+ 1, 0).

1)m=3/2，D(3/2，25/8)

tanADH = AH/DH =( 1+ 1.5)/(25/8)= 4/5

2) When ∠ ADB = 60, an equilateral triangle

√3﹙ 1+m﹚=？ m？ +m﹢？

M =-1+2 √ 3 or-1.

Because B is on the right of A, M =- 1+2 √ 3.

When ∠ ADB = 90, isosceles right triangle

1+m=？ m？ +m﹢？

Similarly, M= 1.

So 1 ≤ m ≤- 1+2 √ 3.

3)AB=2m+2，OC=m+？ ，S 1=(2m+2)(m+？ )/2,

DH intersects BC at F, CT is perpendicular to DH and DG is perpendicular to CB.

FH/CO=BH/BO

(m+ 1)/(2m+ 1)= FH/( 1/2+ 1)

DF=DH-FH=(m？ +2m+ 1)/2-(m？ + 1.5m+0.5)/(2m+ 1)=[(m+ 1)？ (2m+ 1)-(2m？ +3m+ 1)]⊙[2(2m+ 1)]

DF=(m+ 1)m/2

S2=S triangle DCF+S triangle DFB=BO×DF=[(2m+ 1)×(m? +m)/2]/2=S 1

m=2，- 1，- 1/2

So m= 1.

BO=3，CO= 1.5，BC=(3√5)/2

S2= 15/2

So DG=2√5,

So the distance is 2√5.

I made it myself. It should be right.