As shown in the figure, triangle ABC is all equal to triangle ade, if angle BAE = 120 ° angle bad = 40 ° angle BAC =?

As shown in the figure, triangle ABC is all equal to triangle ade, if angle BAE = 120 ° angle bad = 40 ° angle BAC =?

So angle bad + angle DAC = angle DAC + angle CAE so angle bad = angle CAE = 40 angle BAE = angle bad + angle CAE + angle DAC = 80 + angle DAC = 120 angle DAC = 40, so angle BAC = angle bad + angle DAC = 40 + angle DAC = 80 degrees, if you don't understand, please ask me

As shown in the figure, in △ ABC, ∠ B = ∠ C, ∠ bad = 40 ° and ∠ ade = ∠ AED, then the degree of ∠ CDE is______ .

∵∠EDC+∠C=∠AED，∠ADE=∠AED，
∴∠C+∠EDC=∠ADE，
And ? B + ∠ bad =  ADC,
∴∠B+40°=∠C+∠EDC+∠EDC，
∵∠B=∠C．
∴2∠EDC=40°，
∴∠EDC=20°．
So the answer is: 20 degrees

As shown in the figure, if the triangle ABC rotates around its vertex a, the triangle ade is obtained. Is the angle CAE equal to the angle bad? Why

It is equal because ∠ cab = ∠ ead after rotation
If the angle of rotation  cab: ? CAE + ∠ EAB = ∠ cab ∠ bad + ∠ EAB = ∠ cab  CAE = ∠ bad
If the rotation angle ＞∵ cab ? cab = ∠ ead  CAE = ∠ cab + ∠ BAE ∠ bad = ∠ ead + BAE ? CAE = ∠ bad

In triangle ABC, D and E are on AB and AC respectively, and ad ratio DB = 2:1, AE ratio EC = 1:2, then s triangle ade is more than s triangle ABC?

S Delta ADE:S ΔABC=(1/2×2/3AB×1/3AC×sinC):(1/2×AB×AC×sinC)=2/9:1=2:9
∴SΔ ADE:S ΔABC=2:9
Note: s Δ = 1 / 2 × the two sides of the triangle × sin (their included angle)

In the triangle ABC, ad: ab = 1:6, AE: EC = 1:2. If the area of the triangle ade is equal to 1, what is the area of the triangle ABC?

According to the analysis, it can be seen that:
The area of the triangle ade is 1 of the area of the triangle ADC
3，
The area of the triangle ADC is 1 of the area of the triangle ABC
6，
So the area of the triangle ade is 1 of the area of the triangle ABC
3×1
6=1
18，
So the area of the triangle BAC is: 1 ÷ 1
18=18；
A: the area of triangle ABC is 18

In the triangle ABC, the points D and E are on AB and AC respectively. Ad ratio DB is equal to AE ratio EC. Given that the area of triangle ade is equal to 1 and the area of triangle DBC is equal to 12, the area of triangle ABC is calculated

Junior three

As shown in figure a triangle ABC, AE: EC = 1:2, BC: DC = 3:2. The area of triangle ade is 18 square meters. What is the area of triangle ABC?

18 × (1 + 2) × (1 + 2) = 81 square meters

It is known that the triangle ABC is isosceles right triangle, triangle ACD is equilateral triangle, AE vertical CD, AE, BD intersect at O. verification: od = 1 / 2BC This is a problem,

It is proved that: ? ABC is an isosceles right triangle  AB = AC  ABC = 45 ?  ACD is an equilateral triangle ? ad = AC ∠ ADC = ∠ DAC = 60 ° ? AE ⊥ CD ? AED = 90 °? DAE = 30 °  AE = 1 / 2ad, i.e., de = 1 / 2Ac and ? BAC = 90 °∠ DAC = 60 ° ad = ab  ADB = 15 °  ode = 45

The results show that △ ABC is an equilateral triangle, △ abd is an isosceles right triangle, ab = ad, ∠ DAB = 90 ° AE ⊥ BC intersects CD at o at E. verification: od = 2oC

Dear, the title is wrong. We can only prove that BD = 2oC
I can't upload the picture yet. You can draw it yourself
It is proved that because DAC = 90 ° and angle BAC = 60 °,
So ∠ DAC = 150 °
Because Da = AB, ab = AC
So Da = AC ∠ ACD = ∠ ABO = (180-150) / 2 = 15 °
∴∠OCB=∠COE=∠BOE=45°
∴∠DOB=180°-45°-45°=90°
Delta DOB is a right triangle
∠DBO=45+15=60°
So DB = 2ob
OB = OC  DB = 2oC

As shown in the figure, △ ABC, ab = AC, ∠ BAC = 45 °, AE ⊥ BC in E, BD ⊥ AC in D, AE, BD in F, verification: AF = BC, if BF = radical 2, find the length of ab

(1) F is the barycenter of △ ABC; connecting CF and extending the intersection of AB with H, CH ⊥ AB, and ? BAC = 45 ° and  DCF = 45 °; thus, DF = DC and ∵ AFD = ≌≌△ BCD are obtained by the angle angle edge theorem: △ AFD ≌ △ BCD, AF = BC