As shown in the figure, points a, B and E are on the same straight line, △ ABC and △ BDE are equilateral triangles. Ad intersects BC at f and CE intersects BD with AD at g and h respectively

As shown in the figure, points a, B and E are on the same straight line, △ ABC and △ BDE are equilateral triangles. Ad intersects BC at f and CE intersects BD with AD at g and h respectively

① Δ abd ≌ Δ Abe,
② Δ BFD ≌ Δ Bge (part of the base also coincides),
③ Δ Abf ≌ Δ CBG (same as ②)
It's not easy to type,

It is known that in △ ABC, ad is the high line on BC side, CE is the center line on AB edge, DG bisection ∠ CDE, DC = AE, Verification: CG=EG Proof: ∵ ad ⊥ BC ∴∠ADB=90° ∵ CE is the center line on the edge of ab ﹤ e is the midpoint of ab ∴DE=______ (the center line on the hypotenuse of a right triangle is half of the hypotenuse) And ∵ AE = 1 2AB ∴AE=DE ∵AE=CD ∴DE=CD That is, △ DCE is______ triangle ∵ DG bisection ∵ CDE ∴CG=EG（______）

Proof: ∵ ad ⊥ BC,
∴∠ADB=90°，
∵ CE is the center line on the edge of ab,
The midpoint of E is ab,
∴DE=1
2Ab (the center line on the hypotenuse of a right triangle is equal to half of the hypotenuse),
And ∵ AE = 1
2AB，
∴AE=DE，
∵AE=CD，
∴DE=CD，
That is, △ DCE is an isosceles triangle,
∵ DG bisection ∠ CDE,
ν CG = eg (isosceles triangle with three lines in one)
So the answer is: 1
2Ab; isosceles; isosceles triangles

B is a point on the line ad, △ ABC and △ BDE are equilateral triangles The extension line connecting CE and extending ad is at point F, and the circumscribed circle of △ ABC crosses CF at point M Verification: AC 2 = cm times CF

If MB is connected, then abmc four points are in a circle,  CMB + ∠ a = 180 ° = = = > ∠ CMB = 180-60 = 120 °
And: ∠ CBE = 180 - ∠ ABC = 120 °, FCB = ∠ MCB
∴△CBF∽△CMB
∴CF/BC=BC/CM===>CM*CF=BC²=AC²

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = BC, ad bisection ∠ cab intersects BC at point D, de ⊥ AB, the foot perpendicularly is e, and ab = 6cm, then the circumference of △ DEB is () A. 4cm B. 6cm C. 8cm D. 10cm

∵ ad bisection ∠ cab intersects BC at point D
∴∠CAD=∠EAD
∵DE⊥AB
∴∠AED=∠C=90
∵AD=AD
∴△ACD≌△AED．（AAS）
∴AC=AE，CD=DE
∵∠C=90°，AC=BC
∴∠B=45°
∴DE=BE
∵AC=BC，AB=6cm，
ν 2bc2 = AB2, that is BC=
AB2
2=
Sixty-two
2=3
2，
∴BE=AB-AE=AB-AC=6-3
2，
∴BC+BE=3
2+6-3
2=6cm，
The circumference of DEB = de + DB + be = BC + be = 6 (CM)
Another method: after proving the congruence of triangles,
∴AC=AE，CD=DE．
∵AC=BC，
∴BC=AE．
The circumference of △ DEB = DB + de + EB = DB + CD + EB = CB + be = AE + be = 6cm
Therefore, B

In the triangle ABC of the right figure, DC = 2bd, CE = 3aE, the area of shadow part is 20 square centimeter, and the area of triangle ABC is______ Centimeter

According to "DC = 2bd, CE = 3aE", s △ ade = 1
4S△ADC，S△ADC=2
3S△ABC，S△ADE=1
6S△ABC，
So the area of the triangle ABC is 20 ÷ 1
6 = 120 (square centimeter),
A: the area of triangle ABC is 120 square centimeters
So the answer is: 120 square centimeter

In the triangle ABC, DC = 2bd, CE = 3aE, the area of shadow part is 20 square centimeter I have calculated the area of the triangle CDE, which is 60 square centimeters Shadow is ade

Based on similar knowledge
The area of the triangle CDE is 60
So the area of the triangle ADC is 80
So the area of the triangle abd is 40
So the area of the triangle ABC is 120 square centimeters

In the triangle ABC of the right figure, DC = 2bd, CE = 3aE, the area of shadow part is 20 square centimeter, and the area of triangle ABC is______ Centimeter

According to "DC = 2bd, CE = 3aE", s △ ade = 1
4S△ADC，S△ADC=2
3S△ABC，S△ADE=1
6S△ABC，
So the area of the triangle ABC is 20 ÷ 1
6 = 120 (square centimeter),
A: the area of triangle ABC is 120 square centimeters
So the answer is: 120 square centimeter

As shown in the figure, in the triangle ABC, angle B = angle c, D, e, f are on AB, BC, AC respectively, and BD = CE, angle def = angle B Confirmation: ed = EF

∵∠DEF=∠B.
∴∠BDE=180°-∠B-∠BED=180°-∠DEF-∠BED.
And ∠ CEF = 180 ° - ∠ def - ∠ bed
∴∠BDE=∠CEF.
And BD = CE, ∠ B = ∠ C
∴⊿DBE≌⊿ECF(ASA),ED=EF.

As shown in the figure, in the triangle ABC, ab = AC, BD = CE, angle 1 = angle B. It is proved that the triangle DEF is an isosceles triangle (the figure is a little deformed, in the triangle ABC, D is on the AB side) (in the triangle ABC, D is on the AB side, f is on the AC side.. but the position of F is lower than the point below D on the left side. Angle 1 is the angle formed by connecting the points of D and F on BC, which is above the point e.. It may be a little difficult to understand.. I can't think of a topic for half a day to tell us what the use of AB = AC is. After all, only the triangle DBE and the triangle FEC are required to be congruent. BD = CE is enough, But what's the use of AB = AC and angle 1 = angle B? Hope the top mathematicians can draw this graph and find it out.. the chapter we learned is isosceles triangle. Other things that we haven't learned in the previous courses don't appear in the solution!

AB = AC tells us that ∠ B = ∠ C proof: ∵ AB = AC  B =  C ? B = ∠ 1 and ∠ B + ∠ BDE + ∠ DEB = 180 °∠ DEB + ∠ 1 + ∠ FEC = 180 °∠ BDE = ∠ FEC in △ BDE and △ CEF: ∠ BDE = ∠ FEC

In the triangle ABC, AB two AC, points D, e, f are on the edges AB, BC, AC respectively, and BD two CE, angle def = angle B. is there any triangle congruent with the triangle BDE in the graph?

A:
BD=CE,∠B=∠DE‖BC
So: ∠ B = ∠ def = ∠ efc
So: BD ‖ EF
So: the quadrilateral bfed is a parallelogram
So: △ BFE ≌ △ BDE
Because: ∠ B = ∠ C
Because: ∠ Dec = ∠ def + ∠ FEC
There are ∠ Dec = ∠ B + ∠ BDE, ∠ def = ∠ B
So ∠ FEC = ∠ BDE,
So △ BDE is equal to △ CEF
Please click the "select as satisfied" button below
If you have any other questions, please send or click to me for help. The answer is not easy, please understand, thank you!