As shown in the figure, in RT △ ABC, ab = AC, ∠ a = 90 °, point D is any point of BC, DF ⊥ AB is in F, de ⊥ AC is in E, M is the midpoint of BC. Try to judge what shape △ MEF is and prove your conclusion

As shown in the figure, in RT △ ABC, ab = AC, ∠ a = 90 °, point D is any point of BC, DF ⊥ AB is in F, de ⊥ AC is in E, M is the midpoint of BC. Try to judge what shape △ MEF is and prove your conclusion

The proof is as follows: connect am, ∵ m is the midpoint of BC, ∵ BAC = 90 °, ab = AC,  am = 12bc = BM, am bisection  BAC. ? MAC = ∵ mAb = 12 ∠ BAC = 45 °. ∵ ab ⊥ AC, de ⊥ AC, DF ⊥ ab, ∵ AB, DF ⊥ AC, ? BAC = 90 °, quadrilateral dfae

In RT △ ABC, ∠ BAC = 90 °, ab = AC, D is the midpoint of BC, AE = BF It is proved that △ DEF is an isosceles right triangle

Proof: connect ad,
∵ RT △ ABC, ∵ BAC = 90 °, ab = AC,
∴∠B=∠C=45°．
∵AB=AC，DB=CD，
∴∠DAE=∠BAD=45°．
∴∠BAD=∠B=45°．
∴AD=BD，∠ADB=90°．
∵AE=BF，∠DAE=∠B=45°，AD=BD，
∴△DAE≌△DBF（SAS）．
∴DE=DF，∠ADE=∠BDF．
∵∠BDF+∠ADF=∠ADB=90°，
∴∠ADE+∠ADF=90°．
The △ DEF is an isosceles right triangle

Ad parallel BC, angle ABC = angle DCB, ab = DC, AE = DF, connection BF, CE, BF = CE The congruent triangle of knowledge in the second year of junior high school

Because AB = DC, AE = DF
So EF is the midpoint
So CF = be
Should be ad parallel to BC angle ABC = angle DCB
So angle EBC = angle FCB
FC = EB expected
Angle EBC = angle FCB
CB=BC
So triangle FCB congruence and triangle EBC
So BF = CE

In the triangle ABC, BD = DC (D is the midpoint of BC), BF intersects ad, AC at e and f (f points on edge AC). If AF = EF, it is proved that be = AC

Extend ed to H point so that ED = DH
It is easy to prove that bed and CHD are congruent, and the angle bed = DHC be = ch
Because AF = EF, the angle EAF = AEF
Angle AEF = bed
The obtained angle is DHC = EAF
So the triangle AHC is also an isosceles triangle
AC=CH=BE

In the triangle ABC BC, there is d in the middle and between F AD and BF. Under the known condition of E, BD = DC, be = AC, AF = EF

Extend ad to point G such that DG = da
Because DG = Da, DB = DC,
Therefore, ABGC is a parallelogram;
AC ‖ BG, AC = BG
Because, AC ‖ BG,
Therefore, ∠ FAE = ∠ AGB
Because be = AC = BG,
Therefore, ∠ AGB = ∠ Berg
Because, ∠ FAE = ∠ AGB = ∠ Berg = ∠ FEA,
Therefore, AF = EF

As shown in the figure, in the triangle ABC, BD = DC, BF intersects ad, AC and E, F, AF = EF. It is proved that be = AC

prove:
Intercept DM = ad on the extension line of AD and connect BM
∵BD=DC,∠BDM=∠CDA,DM=AD
∴⊿BDM≌⊿CDA（SAS）
∴AC=BM,∠CAD=∠M
∵AF=EF
∴∠EAF=∠FEA
∵∠BEM=∠FEA
∠EAF（∠CAD）=∠M
∴∠BEM=∠M
∴BE=BM
∴BE=AC

In this paper, we prove that the vertex of △ on the intersection line of ABC is an isometric point

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Δ ABC is an equilateral triangle
AB=BC=CA
AB=AF+BF=BD+CD=CE+AE
∵AF=BD=CE
∴BF=CD=AE
∠ a = ∠ B = ∠ C = 90 degrees
So the triangle AEF, BDF, CED are congruent
That is, there is a corresponding edge EF = FD = De
That is, △ DEF is an isosceles triangle

As shown in the figure, in △ ABC, ab = AC, CE, BD are high. Try to prove CE = BD I'm sorry I don't have a picture, but it's really urgent,

prove:
∵AB=AC
∴∠ABC=∠ACB
And ∵ CE and BD are high
∴∠EBC=∠DCB
In ABC
Brace ∠ EBC = ∠ DCB (proved)
Bc-bc (common side)
∠ ABC = ∠ ACB (proved)
 BCE is equal to all BCD (a.s.a)
/ / CE = BD (the corresponding sides of an congruent triangle are equal)
Is this the picture?
Please adopt it

As shown in the figure, △ ABC and △ ade are isosceles right triangles with common vertices （1）BD=CE； （2）∠1=∠2．

It is proved that: (1) ∵ ABC and △ AED are isosceles right angle triangles,