M is the center of gravity of the triangle ABC, and the extension of the BM intersects AC at the poi

Do the extension cable of the AM to cross the BC and E to connect DE

Because M is the center of gravity.

So de ab=1 2

de/ab=dm/mb=1/2

bm=2md

Proof: 1) dm cn, cpm=90°, therefore: 2+ 3=90°;

Again: 1+ 3=90°, therefore: 1= 2;

Again: 2+ 4=90°, so 3= 4;

Again: bc=cd, so: NBC mcd; Therefore, nb=mc;

o is the center of gravity of the square ABCD, so: ob=oc, nbo= mco=45°;

Therefore: triangle onb omc, therefore: om=on;

2) BOC=90°, i.e. BOM+ MOC=90°, OMB OMC, so ON= MOC; BOM+ NOB=90°, i.e.: NOM=90°;

i.e.: om on.

Note: 1= CNB, 2= DMC, 3= NCB, 4= are the intersection of CN and DM.

Proof: The parallel lines of AD through the two points b and c respectively cross the extension lines of cf and be at the two points m and n respectively. Then:

The quadrilateral MBCN is a parallelogram.

Let cm and bn be handed over at point o. By mb ao cn, the empty grinding wheel: of fm=oa bm, oe en=oa cn

And BM=CN

The letter is: of fm=oe en

So: mn ef

And mn bc

So: EF BC,9,As shown in the figure,In the triangle ABC,D is the midpoint of BC,M is a point on AD,BM,The extension lines of CM,CM intersect AC,AB in F,E

Divide line segments with parallel lines and equilateral line segments with parallel lines.

Proof: respectively pass the point m, n as MH parallel AC cross AB to H, NG parallel AC cross AB to G, cross Am to P

So gp de=ap ae=pn ef

mh/ac=bm/bc=bh/ab

mn/nc=hg/ag

bm/mn=bh/hg

Because bm=mn=nc

So ag=hg=1 2ah

hm/ac=1/3

So GP HM=1 2

So gp=1 6ac

So gp pn=1 3

So de ef=1 3

So ef=3de

Proof: pass the point E to do EF BC and pass AC to point F, you can get EFM DCM, EF=1 4BC; Because ae = 1 4ab, f is the midpoint of am, i.e., fm = 1 2am = 1 2cm; From the similarity of triangles, ef=1 2cd; So, cd=1 2bc is bc=2cd.

Make a line finger band diagram: draw an isosceles triangle ABC, make the midline Am on BC, take the points E and E' on AB and AC respectively, connect the reed junctions E and E', extend AB to point F, connect points F and point E', extend AM to point O, connect BO, take point Q on BM, connect QO (QO should be perpendicular to EF).

Solution: Do MH BC over M to H

Because hm bc and abc are equilateral triangles.

So ah=am

bh=cm, ahm= abc= acb=60°, so bhm= mcn

bm=mn again

So bhm mcn

So hm=cn

hm=am again

So am=cn