As shown in the figure, in △ ABC, AC = BC, ∠ ACB = 90 °, D is a point in △ ABC, ∠ bad = 15 °, ad = AC, CE ⊥ ad in E, and CE = 5 (1) Find the length of BC; (2) Confirmation: BD = CD

As shown in the figure, in △ ABC, AC = BC, ∠ ACB = 90 °, D is a point in △ ABC, ∠ bad = 15 °, ad = AC, CE ⊥ ad in E, and CE = 5 (1) Find the length of BC; (2) Confirmation: BD = CD

(1) In △ ABC,
∵AC=BC，∠ACB=90°，
∴∠BAC=45°，
∵∠BAD=15°，
∴∠CAD=30°，
∵CE⊥AD，CE=5，
∴AC=10，
∴BC=10；
(2) It is proved that if D is used as DF ⊥ BC in F
In △ ADC, ∠ CAD = 30 ° and ad = AC,
∴∠ACD=75°，
∵∠ACB=90°，
∴∠FCD=15°，
In △ ace, ∠ CAE = 30 ° CE ⊥ ad,
∴∠ACE=60°，
∴∠ECD=∠ACD-∠ACE=15°，
∴∠ECD=∠FCD，
∴DF=DE．
∵ in RT △ DCE and RT △ DCF,
DC=DC
DE=DF. ，
∴Rt△DCE≌Rt△DCF（HL），
∴CF=CE=5，
∵BC=10，
∴BF=BC-CF=5，
∴BF=FC，
∵DF⊥BC，
∴BD=CD．

As shown in the figure, in △ ABC, ab = AC, ad is high, and E is on AD （1）BD=CD； （2）△BDE≌△CDE； （3）BE=CE．

It is proved that: (1) in △ ABC, ab = AC, ad is high,
ν BD = CD (the height on the bottom edge of an isosceles triangle coincides with the center line on the bottom edge);
(2) ∵ ad is high,
∴∠EDB=∠EDC，
In △ BDE and △ CDE,
ED＝ED
∠EDB＝∠EDC
BD＝CD ，
∴△BDE≌△CDE（SAS）；
（3）∵△BDE≌△CDE，
∴EB=CE．

As shown in the figure, in the triangle ABC, ad is its angular bisector, and BD = CD, De is vertical AB, DF is vertical AC, and perpendicular foot is e F

It is proved that ad is its angular bisector and BD = CD
According to the inverse theorem of three lines in one, ab = AC
Combined equilateral equal angle B = angle c
In RT △ bed and RT △ DFC
Angle B= angle C BD=CD
So △ bed ≌ △ DFC
So be = FC

As shown in Figure 12-3 1-3, in the triangle ABC, ab = AC, BD = BC, ad = de = EB, calculate the degree of ∠ a

AB=AC,
∠ABC=∠C=∠2+∠3
BD=BC
∠C=∠BDC==∠2+∠A=∠2+∠3
∠A=∠3
AD=DE=EB
∠A=∠1；∠2=∠BDE；∠1=2∠2
∠A=∠3=2∠2；∠ABC=∠C=∠2+∠3=3∠2
∠ABC+∠C+∠A=5∠2=180
∠2=36
∠A=2∠2=72
∠A=72

As shown in the figure, in △ ABC, ab = AC, ad = de = EB, BD = BC, try to find the degree of ∠ a

Let ∠ a = x, then
∵AD=DE，∴∠AED=∠A=x；
∵DE=BE，∴∠EDB=∠EBD=1
2x；
And ∵ BD = BC,
∴∠C=∠BDC=∠A+∠EBD=1.5x；
∵AB=AC，∴∠ABC=∠C=1.5x；
In △ ABC, ∠ a + ∠ ABC + ∠ ACB = 4x = 180 °,
∴∠A=x=45°．
So the answer is: 45 degrees

It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°

Let ∠ EBD = x °,
∵BE=DE，
∴∠EDB=∠EBD=x°，
∴∠AED=∠EBD+∠EDB=2x°，
∵AD=DE，
∴∠A=∠AED=2x°，
∴∠BDC=∠A+∠ABD=3x°，
∵BD=BC，
∴∠C=∠BDC=3x°，
∵AB=AC，
∴∠ABC=∠C=3x°，
∵∠A+∠ABC+∠C=180°，
∴2x+3x+3x=180，
The solution is: x = 22.5,
∴∠A=2x°=45°．
Therefore, C

As shown in the figure, in △ ABC, D and E are points on edge AC and BC respectively. If △ ADB ≌ △ EDB ≌ △ EDC, then the degree of ∠ C is () A. 30° B. 45° C. 50° D. 60°

∵△ADB≌△EDB≌△EDC，
∴∠A=∠DEB=∠DEC，∠ADB=∠BDE=∠EDC，
∵∠DEB+∠DEC=180°，∠ADB+∠BDE+EDC=180°，
∴∠DEC=90°，∠EDC=60°，
∴∠C=180°-∠DEC-∠EDC，
=180°-90°-60°=30°．
Therefore, a

In the triangle ABC, the angle c = 2, the angle B, ad is the bisector of the triangle ABC, the angle EDB = the angle B, and it is proved that ab = AC + CD The figure shows that point D is on CB and E is on AB, connecting AD and ed

∠AED=∠EdB+∠B
∠EDB=∠B
∠AED=2∠B,∠C=2∠B
∠AED=∠C,∠CAD=∠DAB
AD=AD
△ACD≌△ADE
AC=AE,
∠EDB=∠B
DE=BE=CD
AB=AE+BE=AC+CD

It is known that in △ ABC, AC = BC, ad bisection ∠ BAC intersects BC at D, point E is the point above AB, and ∠ EDB = ∠ B. There are two conclusions: ① AB = AD + CD; ② AB = AC + CD (1) As shown in Figure 1, if ∠ C = 90 °, the conclusion is drawn______ Yes, and prove your conclusion (2) As shown in Figure 2, if ∠ C = 100 °, the conclusion is drawn______ Yes, and prove your conclusion

（1）∵∠C=90°，AC=BC，
∴∠B=45°，
∵∠EDB=∠B，
∴∠DEB=90°，BE=DE，
∴DE⊥AB，
∵ ad bisection ∵ BAC,
∴DE=DC，
In RT △ ACD and RT △ AED
AD＝AD
DC＝DE ，
∴Rt△ACD≌Rt△AED（HL），
∴AC=AE，
ν AB = AE + be = AC + CD, so ② is correct;
（2）∵∠C=100°，AC=BC，
∴∠B=∠CAB=40°，
∵∠EDB=∠B，
∴∠DEB=100°，BE=DE，
∴∠AED=80°，
∵ ad bisection ∵ BAC,
∴∠DAE=20°，
∴∠ADE=180°-80°-20°=80°，
∴AD=AE，
Through point D, DF ⊥ AC at point F, DH ⊥ AB at point H,
∴DF=DH，
Easy to obtain △ CDF ≌ △ EDH,
∴CD=DE，
∴CD=BE，
ν AB = AE + be = AD + CD, so ① is correct
So the answer is: 1

As shown in the figure, △ ABC is an equilateral triangle, and points D, e and F are points on the segments AB, BC and Ca respectively, (1) If ad = be = CF, is △ def an equilateral triangle? Try to prove your conclusion; (2) If △ DEF is an equilateral triangle, is ad = be = CF true? Try to prove your conclusion

(1) Delta DEF is an equilateral triangle
The proof is as follows:
∵ △ ABC is an equilateral triangle,
∴∠A=∠B=∠C，AB=BC=CA，
And ∵ ad = be = CF,
ν DB = EC = FA, (2 points)
≌ △ ADF ≌ △ bed ≌ △ CFE, (3 points)
/ / DF = de = EF, that is, △ DEF is an equilateral triangle; (4 points)
(2) Ad = be = CF holds
The proof is as follows:
As shown in the figure, ∵ △ DEF is an equilateral triangle,
∴DE=EF=FD，∠FDE=∠DEF=∠EFD=60°，
∴∠1+∠2=120°，
ABC is an equilateral triangle,
∴∠A=∠B=∠C=60°，
∴∠2+∠3=120°，
Ψ 1 = ∠ 3, (6 points)
Similarly, ∠ 3 = ∠ 4,
≌ △ ADF ≌ △ bed ≌ △ CFE, (7 points)
/ / ad = be = cf. (8 points)