As shown in the figure, in △ ABC, point D is on BC, ∠ DAC = ∠ ABC, BD = 4, DC = 5, de ‖ AC, where AB is intersected with E, then De is de=___

Idea: using similar triangles, the corresponding sides are proportional, because ∠ DAC = ∠ ABC, and ∠ C are common angles. So △ ABC and △ DAC are similar. According to the proportion of corresponding sides, BC: AC = AC: DCAC ^ 2 = BC * DC = 45ac = 3 √ 5. Because if De is parallel to AC, then △ ABC and △ EBD are similar. According to the proportion of corresponding sides, de: AC = BD: BCDE

As shown in the figure △ ABC is an equilateral triangle, extend AC to D, make an equilateral △ BDE with BD as the edge, connect AE, and prove that ad = AE + AC

prove:
∵∠ABC=∠EBD=60°
∠ABE=∠ABC-∠EBC
∠CBD=∠EBD-∠EBC
∴∠ABE=∠CBD
And ∵ AB = CB, be = BD
∴△ABE≌△CBD
∴AE=CD
∵AD=AC+CD
∴AD=AC+AE

As shown in the figure, in △ ABC, point E is on BC and point D is on AE. It is known that ∠ abd = ∠ ACD, ∠ BDE = ∠ CDE. Verification: BD = CD

It is proved that: ∵ ADB = 180 ° - ∠ BDE, ∵ ADC = 180 ° - ∠ CDE,
∴∠ADB=∠ADC．
∵ in △ ADB and △ ADC,
∠ABD=∠ACD，AD=AD，∠ADB=∠ADC，
∴△ADB≌△ADC．
∴BD=CD．

As shown in the figure, in the triangle ABC. D is on BC, e is on AB, ab = AC, ad = AE, given the angle DAC = 45, what degree is the angle EDB?

The answer is 22.5 degrees
In the isosceles triangle ade, the angle ade = (180 angle AED) / 2
In the isosceles triangle ABC, the angle ABC = (180-ead-45) / 2
In triangle EDB, the outer angle AED = angle EBD + angle EDB
EDB + EDA + ADC = 180
The answer can be found from the above

It is known that in △ ABC, the intersection of the vertical lines of a and BC is D, and then the vertical line angles AB and e of D are made, ad = BD, ∠ DAC = ∠ BDE 1) Verification: △ ABC ∽ EAD 2) If ad ⊥ BC, ∠ C = 60 ° and CD = 1, find the length of de (the result is accurate to 0.01M), please write the result····

1. Because ad = BD and ad ⊥ BD, so ∠ B = ∠ BDE = ∠ DAC = 45 °. A = 90 °. ABC is an isosceles right triangle, △ ead is also an isosceles right triangle, so △ ABC ∽ EAD. 2

As shown in the figure, in triangle ABC, AB is equal to AC, D is a point on the edge of BC, angle B is 30 ° and angle DAB is 45 °. It is proved that DC is equal to ab

The ∵ AB = AC, ∵ B = AC, ∵ B = C = 30, ∵ we \ \\\\\\\\\\\\\\it's a good idea

As shown in the figure, in right angle trapezoid ABCD, ∠ ABC = 90 °, ad ‖ BC, ab = BC, e is the midpoint of AB, CE ⊥ BD (1) Results: be = ad; (2) Verification: AC is the vertical bisector of ED; (3) Is △ DBC an isosceles triangle? And explain the reason

(1) It is proved that ∵ ABC = 90 ° BD ⊥ EC,
∴∠1+∠3=90°，∠2+∠3=90°，
∴∠1=∠2，
In △ bad and △ CBE,
∠2＝∠1
BA＝CB
∠BAD＝∠CBE＝90° ，
∴△BAD≌△CBE（ASA），
∴AD=BE．
(2) It is proved that ∵ e is the midpoint of ab,
∴EB=EA，
∵AD=BE，
∴AE=AD，
∵AD∥BC，
∴∠7=∠ACB=45°，
∵∠6=45°，
∴∠6=∠7，
And ∵ ad = AE,
⊥ De, and EM = DM,
AC is the vertical bisector of ED;
(3) Δ DBC is an isosceles triangle (CD = BD)
The reasons are as follows.
∵ from (2), CD = CE, from (1), CE = BD,
∴CD=BD．
The △ DBC is an isosceles triangle

As shown in the figure, in △ ABC, point D is on BC, and ∠ ABC = ∠ ACB, ∠ ADC = ∠ DAC, ∠ DAB = 24 ° to find the degree of ∠ ABC; and answer: which triangles in the graph are obtuse angle triangles?

According to the relation of exterior angle of triangle, we get ∠ ADC = ∠ ABC + ∠ DAB. From the sum theorem of inner angles of triangle, we can get: ∠ ADC + ∠ DAC + ∠ ACB = 180 ° and substitute ∠ DAC = ∠ ADC, ∠ ACB = ∠ ABC into the above formula, SO 2 ∠ ADC + ∠ ABC = 180 ° is obtained by substituting ∠ ADC = ∠ ABC + DAB into the above formula: 2 ∠ ABC + 2 ∠ D

In △ ABC, ∠ C = 90 ° and the perpendicular lines of AB intersect BC and D respectively, AB and E. ① if ∠ CAD = 10 °, calculate the degree of ∠ B; ② if ∠ DAC: ∠ DAB = 1:1

first
Angle B equals 40 degrees
because
∠C=90° BE = AE
therefore
∠B=∠B+∠B+10°+ 90°=180°
So ∠ B = 40 degrees
second
because
∠C=90°∠DAC：∠DAB=1
therefore
AE = be because in a right triangle, the opposite side of 30 ° is half of the hypotenuse
So, it turns out that,
∠B=30°
Suppose ∠ DAC = ∠ DAB = ∠ H
sin∠B=AC/AB
BE=AE=2/1AB
SIN30=0.5
If there is something wrong, please advise

As shown in the figure, ∠ ABC = 42 ° in △ ABC, D is a point on the edge of BC, DC = AB, and ∠ DAB = 27 ° (1) Delta ABC is______ triangle; (2) Prove your conclusion

(1) (2) isosceles. (2) as shown in the figure: turn △ ADB along ad to get △ ade, and △ ADB ≌ △ ade,  5 = ∠ bad = 27 °, 7 = ∠ B = 42 °, BD = De,  2 + ∠ DAB = 69 °, ν 6 = 111 ° - ∠ 2 = 42 ° = ∠ B = ∠ 7, ∵ MD = me, ∵ DC = ab

As shown in the figure, in △ ABC, point D is on BC, ∠ DAC = ∠ ABC, BD = 4, DC = 5, de ‖ AC, where AB is intersected with E, then De is de=___