As shown in the figure, in △ ABC, ∠ ACB = 90 °, points D and E are on AB, ad = AC, be = BC. Try to judge whether the size of ∠ DCE is related to the degree of ∠ B. if so, ask for the relationship between them; if not, please determine the degree and explain the reason

As shown in the figure, in △ ABC, ∠ ACB = 90 °, points D and E are on AB, ad = AC, be = BC. Try to judge whether the size of ∠ DCE is related to the degree of ∠ B. if so, ask for the relationship between them; if not, please determine the degree and explain the reason

The degree of DCE and B is independent,
The reason is: ∵ ACB = 90 °,
∴∠B+∠A=90°，
∵AD═AC，BE=BC，
∴∠ADC=∠ACD=1
2（180°-∠A），∠BEC=∠BCE=1
2（180°-∠B），
∴∠DCE=180°-∠ADC-∠BEC）
=180°-1
2（180°-∠A）-1
2（180°-∠B）
=1
2∠A+1
2∠B
=1
2×90°
=45°，
That is, DCE is always equal to 45 degrees

As shown in the figure, in the RT triangle ABC, ∠ ABC = 90 °, CD is high, CE is the bisector of ACB, ∠ a = 20 ° and the degree of ∠ DCE is calculated

∵∠A=20°,∠ACB=90°
∴∠B=70°
∵CD⊥AB
∴∠BAD=20°
∵ CE bisection ∵ ACB
∴∠BCE=45°
∴∠DCE=45-20=25°

As shown in the figure, in △ ABC, ∠ C is a right angle. The high Cd on AB and the central line CE exactly divide ∠ ACB. If AB = 20, what are the two acute angles of △ ABC and the values of AD, de and EB?

∵ C is a right angle, CD and CE divide  ACB into three equal parts,
∴∠ACD=∠DCE=∠ECB=1
3×90°=30°，
∵ CD is high,
∴∠A=90°-∠ACD=90°-30°=60°，
∵ CE is the midline,
∴CE=AE=EB=1
2AB=1
2×20=10，
∴∠B=∠ECB=30°，
∴AC=1
2AB=1
2×20=10，
AD=1
2AC=1
2×10=5，
DE=AE=AD=10-5=5．
To sum up: ∠ a = 60 °, B = 30 °, ad = 5, de = 5, EB = 10

As shown in the figure, D, e and F are points on the three sides of △ ABC, and the areas of CE = BF, △ DCE and △ DBF are equal Verification: ad bisection ∠ BAC

It is proved that DN ⊥ AC, DM ⊥ AB are made by D,
DBF = 1
2BF•DM，
The area of △ DCE is 1
2DN•CE，
The area of ∵ △ DCE and △ DBF are equal,
∴1
2BF•DM=1
2DN•CE，
∵CE=BF，
∴DM=DN，
Ψ ad bisection ∠ BAC (points with equal distance to both sides of the corner are on the bisector of the angle)

As shown in the figure, in △ ABC, ∠ ACB = 90 °, B = 60 °, CD, CE are the bisectors of the height and angle of △ ABC respectively. Calculate the degrees of ∠ DCE and ∠ AEC

∵ CE is the angular bisector of ∵ ABC,
∴∠ACE=∠BCE=45°，
In △ ABC, ∠ B = 60 °,
∴∠BCD=30°，
∴∠DCE=∠ECB-∠DCB=45-30=15°，
∠AEC=∠BCE+∠B=105°．

As shown in the figure, ∠ ABC = 90 ° in RT △ ABC. Rotate RT △ ABC clockwise for 60 ° around point C to obtain △ Dec, and point E is on AC Then take the straight line AB as the symmetry axis to make the axisymmetric figure of RT △ ABC △ ABF

It is proved that: (1) rt △ Dec is obtained by the rotation of RT △ ABC around point C by 60 ° and  AC = DC, ﹤ ACD = 60 ° and ∵ ad = DC = AC (1 point) and ∵ RT △ ABF is obtained by turning RT △ ABC along the line of AB by 180 ° to obtain  AC = AF, ﹤ ABF = ﹤ ABC = 90 °  FBC is a flat angle  points F, B

As shown in the figure, ∠ ABC = 90 ° in RT △ ABC. Rotate RT △ ABC clockwise for 60 ° around point C to obtain △ Dec, point E on AC, and then flip RT △ ABC along the line of AB by 180 ° to obtain △ ABF. Connect ad (1) The results show that the quadrilateral AFCD is rhomboid; (2) Connect be and extend intersection ad to g, connect CG, what is quadrilateral ABCG special parallelogram and why?

(1) It is proved that RT △ Dec is obtained by rotating RT △ ABC about point C by 60 degrees,
∴AC=DC，∠ACB=∠ACD=60°，
The △ ACD is an equilateral triangle,
/ / ad = DC = AC, (1 point)
And ∵ RT △ ABF is obtained by turning RT ∵ ABC along the line of AB by 180 degrees,
∴AC=AF，∠ABF=∠ABC=90°，
∵∠ACB=∠ACD=60°，
△ AFC is an equilateral triangle,
/ / AF = FC = AC, (3 points)
∴AD=DC=FC=AF，
The quadrilateral AFCD is rhombic. (4 points)
(2) Quadrilateral ABCG is rectangular. (5 points)
It is proved that: (1) we know that △ ACD, △ AFC are equilateral triangles, △ ACB ≌ △ AFB,
∴∠EDC=∠BAC=1
2 ∠ fac = 30 ° and △ ABC is a right triangle,
∴BC=1
2AC，
∵EC=CB，
∴EC=1
2AC，
Ψ e is the midpoint of AC,
∴DE⊥AC，
/ / AE = EC, (6 points)
∵AG∥BC，
∴∠EAG=∠ECB，∠AGE=∠EBC，
∴△AEG≌△CEB，
ν Ag = BC, (7 points)
The quadrilateral ABCG is a parallelogram,
∵ ABC = 90 °, (8 points)
The quadrilateral ABCG is a rectangle

As shown in the figure, it is known that △ ABC and △ Dec are equilateral triangles, ∠ ACB = ∠ DCE = 60 °, B, C, e are on the same line, connecting BD and AE Confirmation: DF = Ge

BD intersects AC in F, AE crosses CD in G ∵ ? ? ? ? ? ?  ABC and △ Dec are equilateral triangles ? AC = BC, CD = CE, ﹤ ACB = ∠ DCD = 180 ° - ∠ ACB - ∠ DCE = 60 °∠ ace = ∠ ACE + ∠ DCE = 120 ° in △ ace and △ BCD, AC = BC, CD = CE, ∠ ace = 3bcd = 12

In △ ABC, ∠ ACB = 90 °, points D and E are on AB, and ad = AC, BC = be. Find the degree of ∠ DCE

 ad = AC, BC = be,  ACD = ∠ ADC,  BCE = ﹤ BEC,  ACD = (180 ° - ∠ a) ﹤ 2 ①,  BCE = (180 ° - ∠ b) ﹤ 2 ②, ? a + ∠ B = 90 °,  ① + ② - ﹤ DCE, then ﹤ ACD + ∠ BCE - ∠ DCE = 180 ° - (∠ a + ∠ b) ﹤ 2 - ﹥ DCE = 180 ° - 45 ° - ﹤ DCE = 135 ° -

As shown in the figure, in RT △ ABC, ∠ ACB = 90 ° and CD is the height above AB edge. If ad = 8 and BD = 2, find CD

∵ in RT △ ABC, ∵ ACB = 90 ° and CD is the height of AB edge
∴∠BDC=∠ACB=90°
∵∠B=∠B
∴△ABC∽△CBD
∴CD2=AD•BD，
∵AD=8，BD=2，
∴CD=
8×2=4．