Connect ec to prove that ec=be comes first.

Then because ∠ 1=∠2=∠3 and ∠cef is the common angle of triangular electrocardiogram and efc.

So △ecg and △efc are almost ef/ec=ec/eg.

Understand? =ec？ =ef eg

The second question:

(1): branch 1 ∠5=∠6

Because ∠ ADC+∠ 1 = 90, ∠ ADC+∠ 4+90, ∠ 1=∠4.

So ∠2=∠ 1=∠4.

∠5=∠6 and ∠2=∠4△AME∽△NMD.

(2): na=nd because there is a median vertical line.

∠ 1+∠7=∠adc，∠2=∠4

So ∠ 1+∠ 2+∠ 7 = ∠ 4+∠ ADC = 90.

And because ∠ NCA = 90, it is easy to prove that △nab is similar to △nca.

nc/na=na/nb

Take na? = NC nb so ND？ =NC NB

Question 3: I found two ways to help you.

To prove ∠ mbg = 90, just prove it? GBD~？ GMB，

That is, BG 2 = DG× mg = DG (MD+DG)?

bg^2=bd^2+dg^2=dg(md+dg)=dg^2+dg×md

BD^2=DG×MD=EF× 1/2(AD+HD)

And BD 2 = 1/4? BC^2

AD+HD = DC tanc+BD cot∠BHD = 1/2？ BC(tanC+cotC)

EF=CE？ sinC=BCcosC？ sinC？

EF× 1/2？ (AD+HD)=BC？ cosC？ sinC× 1/2× 1/2？ BC(tanC+cotC)= 1

1/4? BC^2？ (sin^2？ (C)+？ cos^2(C))= 1/4？ BC^2

So BD 2 = ef× 1/2 (AD+HD), and the certificate is finished!

Let ab = a and BC = B.

Because AD⊥BC is derived from Pythagorean theorem.

AD =√ (AB^2-BD^2)=√[a^2-(b/2)^2)

Because ad ⊥ BC and be ⊥ AC are derived from the triangle area formula.

*√[a^2-(b/2)^2]

Because BE⊥AC is derived from Pythagorean theorem.

CE =√(BC^2-BE^2)=b^2/(2a)

It is easy to know that △BDH and △BEC are similar from AD ⊥ BD and BE ⊥ EC, so

hd=bd*ce/be=b^2* √[a^2-(b/2)^2]/{4*[a^2-(b/2)^2]}

therefore

ah=ad-hd=(4*a^2-2b^2)* √[a^2-(b/2)^2]/{4*[a^2-(b/2)^2]}

Known as AM=MH

mh=ah/2=(2*a^2-b^2)* √[a^2-(b/2)^2]/{4*[a^2-(b/2)^2]}

Because ef ⊥ BC, be ⊥ EC and DG = ef are derived from the triangle area formula.

dg=ef=be*ce/bc=b^2/(2*a^2)*√[a^2-(b/2)^2]

therefore

md=mh+hd=2*a^2* √[a^2-(b/2)^2]/{4*[a^2-(b/2)^2]}

therefore

MD * DG = B 2/4 = BD 2 (this is the formula for calculating the height on the hypotenuse of a right angle △).

Because BD⊥MG

So △MBG is a right angle △

So BG⊥BM

If you feel that the right angle △ hypotenuse height formula has not been learned, you can pass it.

MD*DG=BD^2

It is proved that △BDM is similar to △GDB.

So ∠MBD=∠BGD.

Because BD⊥MG

So ∠BDM=90 degrees.

So ∠ MBG = ∠ MBD+∠ GBD = ∠ BGD+∠ GBD =180 degrees -∠BDM=90 degrees.

So BG⊥BM