In ABC, AM is the midline and AE is the high line. Proof of AB AC 2 AM BM

The downstairs is well done. Both graphs, but the principle is the same, it can be like this: because ae is high, ae is perpendicular to bc, so ab 2 = ae 2 + be 2, ac 2 = ae 2 + ce 2; So ab2 ac 2=be 2+ce 2+2ae 2; (1) And because am is the midline, so BM=cm, so be 2=(BM-me) 2=(cm-me) 2=cm 2+me 2-2cm*me; (2) In the same way, CE 2=(cm+me) 2=CM2+ME2+2cm*ME; (3) The above (2) and (3) formulas are added together, be 2 + ce 2 = 2cm 2 + 2me 2; (4) Substituting Eq. (4) into Eq. (1) obtains, ab 2 ac 2 =2cm 2+2me 2+2ae 2 =2bm 2+2(me 2+ae 2) =2bm 2+2am 2 proposition is proved.

Proof : AE is parallel to BC, so EAD= cmd

Because d is the midpoint of the line segment AM.

So ad=md, and ade= mdc

So ade mdc

So ae=mc

Because AM is the midline of ABC.

So BM=MC, so AE=BM, because AE is parallel to BC

So the quadrilateral AEBM is a parallelogram.

So be=am, because am=ac

So be=ac

So the quadrilateral EBCA is an isosceles trapezoid.

AM is the midline, i.e., BM=cm

be=bm+me，ec=cm-me=bm-me∵ae⊥bc

In RT ABE.

ae²=ab²-be²=ab²-(bm+me)²…1) In RT ace.

ae²=ac²-ec²=ac²-(bm-me)²…2) In RT AME.

ae²=am²-me²……3）

2ae²=ab²+ac²-2bm²-2me²……4) (3) 2: 2ae = 2am -2me ......5) (4) and (5).

ab²+ac²-2bm²-2me²=2am²-2me²ab²+ac²=2am²+2bm²

i.e. ab + ac = 2 (am + bm).

There is no picture, just follow what I draw. Let em be x, then there is ab 2+ac 2=be 2+ae 2+ec 2+ae 2=2ae 2+(bm-x) 2+(bm+x) 2=2bm 2+2x 2+2ae 2. And x 2 + ae 2 = am 2, so ab 2 + ac 2 = 2 (am 2 + bm 2).

.AM is the midline of ABC.

bm=cm=ce+em

AE is the high line of ABC.

ab2=be2+ae2，ac2=ae2+ce2，am2=me2+ae2

ab2+ac2

be2+ae2+ae2+ce2

bm+me)2+ae2+ae2+(cm-me)2=(bm+me)2+ae2+ae2+(bm-me)2=bm2+2bmme+me2+ae2+ae2+bm2-2bmme+me2

2bm2+2ae2+2me2

2bm2+2(ae2+me2)

2bm2+2ame2

The formula for finding the opposite sides of a triangle with known sides and angles c 2=a 2+b 2-2abxcosc Yu Xuan's theorem.

Let the angles between AM and BC be y, z. respectively

ab²=am²+bm²-2am x bm x cosy

ac²=am²+cm²-2am x cm x cosz=am²+bm²-2am x bm x cosz

Add the two sides. ab²+ac²=am²+bm²-2am x bm x cosy+am²+bm²-2am x bm x cosz

2(am²+bm²)-2am x bm x (cosy+ cosz)

The cosine function value relationship of the complementary angle cos(a)=cos(180-b)=-cos(-b)=-cos(b).

y, z is the complementary angle, 2am x bm x (cosy + cosz) = 0

So ab + ac = 2 (am + bm ).

Extend AM as MD=MA to obtain the parallelogram ABCD

AB +AC = 2(AM +BM) (Parallelogram property theorem: the sum of the squares of the diagonal of the parallelogram is equal to the sum of the squares of the four sides).

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I just had the same topic this morning, hehe.