So angel ·BMG = angel ·ACB

Because of Amy. BD。 MF are all high-speed lines.

So the angle BEM=90 degrees

Angle BDC= Angle MFC=90 degrees.

So MF is parallel to BD.

The quadrilateral MFDG is a rectangle.

So MF=DG

Angle BGM=90 degrees

Because AB=AC

So angle M\EBM= angle ACB.

So angle EBM= angle BMG

Angle BEM= Angle BGM=90 degrees

Because BM=BM

So triangle boundary element and triangle boundary element are congruent (AAS)

So BE=BG

Because BD=BG+DG

So BE+MF=BD

2. It is proved that the intersection e in G is EG vertical BC, and the intersection f in H is FH vertical EG.

So EGC angle = EHF angle =90 degrees.

So FH parallel MG

So angle EFH= angle BCE

Because FM, FN fn are both high lines.

So ENF angle FMC angle =90 degrees.

So angle FMC= angle EGC=90 degrees.

So for example, parallel frequency modulation

So the quadrilateral FHGM is a rectangle.

So FM=HG

Because BC=BE

So angle NEF= angle BCE

So angle NEF= angle EFH

Because EF=EF

So triangle ENF and triangle EHF are congruent (AAS0

So FN=EH

Because EG=EH+HG

So FM+FN=EG

So FN+FM= the perpendicular from point E to BC.

3. solution: the intersection point m is MG, and the vertical d is g.

So the angle BGM= the angle OGM=90 degrees.

Because the distance from m to BD is 3/3 of the root number.

So MG= root number 3/3.

Because ABCD is rectangular.

So OB=OD=OA= 1/2BD.

Angle error =90 degrees

Because AOD angle =60 degrees

So the triangle AOD is a regular triangle.

So the angle ADB=60 degrees.

So MBG angle =30 degrees.

So BG= 1.

BM^2=MG^2+MG^2

So BM=2 times the root number 3/3.

Because BD=4

So OB=2

Because OB=BG+OG

So OG=BG= 1

Because MG=MG

So triangle BMG and triangle OMG are congruent (SAS)

So angle ABD= angle MOB=30 degrees.

BM=OM

So OM=2 times the root number 3/3.

So point M(-2 times the root number 3/3,0)

Extend BG so that BH=BG, the intersection H is HM', the vertical BD is at the intersection of AB and M', the intersection M' is M'N, and the vertical X-axis is at n..

So angle H= angle BGM'=90 degrees.

Angle M'NM=90 degrees

Because the angle M'MN= the angle ABD+ the angle MOB=60 degrees.

So the angle HM'M=30 degrees

So NM= 1/2MM/

M'N= radical sign (mm' 2-Mn 2)

Because angle HBM'= angle MBG

So triangle BHM and triangle BGM are congruent (ASA)

So HM'=MG

BM=BM'= 1/2MM '

So MM/=4 times the root number 3/3.

So NM=2 times the root is 3/3 is better.

M'N=2

ON=NM+OM=4 times the root number 3/3.

So the point M/(-4 times the root number 3/3, -2)

To sum up, the distance from the point M on the straight line AB to BD is the midpoint M(-2 times the root number 3/3,0) or M(-4 times the root number 3/3,2).