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As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are on BC and AC respectively, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles triangle As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are respectively on BC and AC, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles right triangle. Please explain the reasons
Even cm
∵ m is the midpoint of the slant of RT ᙽ M
∴MC=AB/2=MB
∠MCE=45°=∠MBD
CE = BD
∴△MCE≌△MBD
∴ME=MD
△mde equilateral
As shown in the figure, the triangle ABC is an isosceles triangle, ∠ ACB = 90 degrees, the midpoint D of BC is taken as the vertical AB, and the vertical foot is e connected with CE, and the value of sin ∠ ace is calculated
Let be = de = a, then BD = DC = (√ 2) ABC = AC = (√ 2) AB = (√ 2) a, BC ^ 2 = AC ^ 2 = 8A ^ 2Ab = (√ 2) BC = (√ 2) a * (2 √ 2) a = 4aae = 3A. According to the cosine theorem, CE ^ 2 = be ^ 2 + BC ^ 2-2be * BC * CoSb = a ^ 2 + 8A ^ 2-2a * (2
As shown in the figure, △ ABC, D and E are the midpoint of edges BC and ab respectively, and AD and CE intersect at G Verification: GE CE＝GD AD＝1 3．
Proof: connected to ed
∵ D and E are the midpoint of side BC and ab respectively,
∴DE∥AC，DE
AC＝1
2，
∴∠ACG=∠DEG，∠GAC=∠GDE，
∴△ACG∽△DEG．
∴GE
GC＝GD
AG＝DE
AC＝1
2，
∴GE
GE+CG=GD
GD+AG，
∴GE
CE＝GD
AD＝1
3．
As shown in the figure, △ ABC, D and E are the midpoint of edges BC and ab respectively, and AD and CE intersect at G Verification: GE CE＝GD AD＝1 3．
Proof: connected to ed
∵ D and E are the midpoint of side BC and ab respectively,
∴DE∥AC，DE
AC＝1
2，
∴∠ACG=∠DEG，∠GAC=∠GDE，
∴△ACG∽△DEG．
∴GE
GC＝GD
AG＝DE
AC＝1
2，
∴GE
GE+CG=GD
GD+AG，
∴GE
CE＝GD
AD＝1
3．
As shown in the figure, △ ABC, D and E are the midpoint of edges BC and ab respectively, and AD and CE intersect at G Verification: GE CE＝GD AD＝1 3．
Proof: connected to ed
∵ D and E are the midpoint of side BC and ab respectively,
∴DE∥AC，DE
AC＝1
2，
∴∠ACG=∠DEG，∠GAC=∠GDE，
∴△ACG∽△DEG．
∴GE
GC＝GD
AG＝DE
AC＝1
2，
∴GE
GE+CG=GD
GD+AG，
∴GE
CE＝GD
AD＝1
3．
As shown in the figure, △ ABC, D and E are the midpoint of edges BC and ab respectively, and AD and CE intersect at G Verification: GE CE＝GD AD＝1 3．
Proof: connected to ed
∵ D and E are the midpoint of side BC and ab respectively,
∴DE∥AC，DE
AC＝1
2，
∴∠ACG=∠DEG，∠GAC=∠GDE，
∴△ACG∽△DEG．
∴GE
GC＝GD
AG＝DE
AC＝1
2，
∴GE
GE+CG=GD
GD+AG，
∴GE
CE＝GD
AD＝1
3．
As shown in the figure, △ ABC, D and E are the midpoint of edges BC and ab respectively, and AD and CE intersect at G Verification: GE CE＝GD AD＝1 3．
Proof: connected to ed
∵ D and E are the midpoint of side BC and ab respectively,
∴DE∥AC，DE
AC＝1
2，
∴∠ACG=∠DEG，∠GAC=∠GDE，
∴△ACG∽△DEG．
∴GE
GC＝GD
AG＝DE
AC＝1
2，
∴GE
GE+CG=GD
GD+AG，
∴GE
CE＝GD
AD＝1
3．
As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD
It is proved that in RT △ AEC, AF ⊥ EC,
∴AC2=CF•CE．
In RT △ ABC, ad ⊥ BC,
∴AC2=CD•CB．
∴CF•CE=CD•CB．
∴CF
CB＝ CD
CE．
∵∠DCF=∠ECB，
∴△DCF∽△ECB．
∴∠B=∠CFD．
As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD
It is proved that in RT △ AEC, AF ⊥ EC,
∴AC2=CF•CE．
In RT △ ABC, ad ⊥ BC,
∴AC2=CD•CB．
∴CF•CE=CD•CB．
∴CF
CB＝ CD
CE．
∵∠DCF=∠ECB，
∴△DCF∽△ECB．
∴∠B=∠CFD．
As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD
It is proved that in RT △ AEC, AF ⊥ EC,
∴AC2=CF•CE．
In RT △ ABC, ad ⊥ BC,
∴AC2=CD•CB．
∴CF•CE=CD•CB．
∴CF
CB＝ CD
CE．
∵∠DCF=∠ECB，
∴△DCF∽△ECB．
∴∠B=∠CFD．
RELATED INFORMATIONS
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