It is known that in △ ABC, ab = AC, CD ⊥ ad is in D, CD = 1 2BC, D is outside △ ABC, and it is proved that ∠ ACD = ∠ B

It is known that in △ ABC, ab = AC, CD ⊥ ad is in D, CD = 1 2BC, D is outside △ ABC, and it is proved that ∠ ACD = ∠ B

It is proved that the point a is the AE ⊥ BC, and the point B C is the point E,
∴∠AEC=90°．
∵AB=AC，
∴BE=1
2BC．
∵CD=1
2BC，
∴BE=CD．
∵CD⊥AD，
∴∠D=90°．
In RT △ Abe and RT △ ACD
AC＝AC
BE＝CD ，
∴Rt△ABE≌Rt△ACD（HL）
∴∠ACD=∠B．

As shown in the figure, △ Abe and △ ACD are formed by △ ABC turning 180 ° along the sides of AB and AC respectively. If ∠ BAC = 150 °, then the degree of ∠θ is () A. 60° B. 50° C. 40° D. 30°

∵∠BAC=150°
∴∠ABC+∠ACB=30°
∵∠EBA=∠ABC，∠DCA=∠ACB
Ψ EBA + ∠ ABC + ∠ DCA + ∠ ACB = 2 (∠ ABC + ∠ ACB) = 60 °, i.e., ∠ EBC + ∠ DCB = 60 °
∴θ=60°．
Therefore, a

In the RT triangle ABC, if angle ACB = 90 degrees, AE = AC, BD = BC, then angle ACD + angle BCE =?

The angle is ACB = 90 degree, AE = AC, BD = BC, CD, CE, etc, and the CD, CE, ACD, CD, CE, ACD, etc, are ＜ 1, the ＜ DCE is ＜ 2, and the ＜ BCE is ＜ 3. The AEC = 1 + 1 + 2 ＜ BDC = 2 ＜ 2 ＜ 2 ＜ 2 ＜ 2 ＜ BDC = 2 ＜ 2 therange of DCE is: the angle of ACB = 90 degree, AE = AC, BD = BD = BC, the connection of CD, CE, ACD, CD, CE ＜ ACD, CD, CE, ACD is ＜ 1, the ＜ 1, the ＜ DCE is ＜ 2, the ＜ 3 ＜ 3 3 = 45 ° ACD

In △ ABC, ∠ ACB = 90 ° and ∠ ABC = 30 ° respectively, positive △ ACD and positive △ BCE are respectively made with AC and BC as the edges, and AE and BD are connected to intersect at O. it is proved that ∠ AOD = 60 °

It is proved that in positive △ ACD and positive △ BCE, AC = CD, BC = CE, ∠ ACD = ∠ BCE = 60 °,
∴∠ACD+∠ACB=∠BCE+∠ACB，
∴∠ACE=∠DCB，
In △ and △ B,
AC＝CD
∠ACE＝∠DCB
BC＝CE ，
∴△ACE≌△DCB（SAS），
∴∠CAE=∠CDB，
∴∠ODA+∠OAD=∠ODA+∠CAD+∠CAE，
=∠ODA+∠CDB+∠CAD，
=∠CDA+∠CAD，
=120°，
In △ oad, ∠ AOD = 180 ° - (∠ ODA + ∠ OAD) = 180 ° - 120 ° = 60 °,
Therefore: ∠ AOD = 60 °

In the triangle ABC, AC and BC are taken as the edge oriented forms respectively, and the equilateral triangles ACD, BCE, BD and AE intersect at M and connect cm. It is proved that CM bisects the angle DME

prove:
∵ ⊿ ACD and ⊿ BCE are equilateral triangles
∴AC=DC,BC=EC,∠ACD=∠BCE=60º
∴∠ACD+∠ACB=∠BCE+∠ACB
That is ∠ DCB = ∠ ace
∴⊿DCB≌⊿ACE（SAS）
∴BD=AC,S⊿DCB=S⊿ACE
Cm ⊥ BD in M, CN ⊥ AE in n
Then s ⊿ DCB = 1 / 2 cm × BD, s ⊿ ace = 1 / 2 cn × AE
ν cm = CN [or write directly the height on the corresponding side of the congruent triangle without writing the area]
ν cm bisection ∠ DME [the point with equal distance to both sides of the corner is on the bisector of the angle]

Take the sides AC and BC of the triangle ABC as the sides to make the equilateral triangle ACD and the triangle BCE respectively, connect AE, BD and intersect at the point O, and prove that BD = AE

It can be proved by using the congruence of a triangle
DC=AC
∠DCB =∠ACE
BC=EC
△DBC≌△AEC（SAS）
So it can be proved that BD = AE

In the RT triangle ABC, if ACB = 90, AC = AE, BD = BC, then ACD + BCE=____ (on the hypotenuse AB, the order of points is a, D, e, b)

In two isosceles triangles
AEC=(180-A)/2;
BDC=(180-B)/2;
So DCE = 45;
So ACD + BCE = 90-dce = 45

As shown in the figure, △ ABC, point D is on edge AB, and ∠ ACD = ∠ ABC is satisfied. If AC = 2, ad = 1, then the length of DB is () A. 1 B. 2 C. 3 D. 4

∵∠ACD=∠ABC，∠A=∠A，
∴△ABC∽△ACD，
∴AB
AC＝AC
AD，
∵AC=2，AD=1，
∴1+DB
2＝2
1，
DB = 3
Therefore, C

It is known that the triangle ABC and the triangle ECD are isosceles right triangles. The angle ACB = angle DCE = 90, and D is a point on the edge of ab (1) Triangle ace congruent triangle BCD (2)AD^2+AE^2=DE^2

Because AC = CB, CD = CE
Angle ACE + angle ACD = 90
Angle DCB + angle ACD = 90
So angle ace = angle DCB
Is it possible to prove congruence if two sides are equal and the angles on both sides are equal
Because it's congruent, so
Angle CAE = angle ABC = 45
Because it is isosceles right angle, so the angle cab = 45
So angle BAE = angle CAE + angle cab = 90
So the AED triangle is a right triangle
So according to Pythagorean theorem, ad ^ 2 + AE ^ 2 = de ^ 2

As shown in the figure, triangle ABC is similar to triangle ACD and ad = 3, BD = 2, and the similarity ratio of triangle ACD and triangle ABC is------

Because the triangle ABC is similar to the triangle ACD, AB / AC = AC / ad AC ^ 2 = AB * AD1) d is in the segment AB, ad = 3 AB = AD + DB = 5 AC ^ 2 = 15 AC = √ 15, the similarity ratio of triangle ACD and triangle ABC = AC / AB = √ 15 / 52) d is outside the segment AB, ad = 3 AB = ad-db = 1 AC ^ 2 = 3 AC = √ 3, triangle ACD and three