As shown in the figure, it is known that in △ ABC, ab = AC, D is a point on the CB extension line, ∠ ADB = 60 °, e is a point on ad, and de = dB
It is proved that in △ abd and △ ACF, ab = AC, ab = AC, ABC = ACB, abd = ACF, ab = AC, abd = acfbd = CF, abd = SAS, ad = AF, and ADB = 60 °, ADF is an equilateral triangle, ad = DF, ad = AF
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- 1. As shown in the figure, it is known that in △ ABC, ab = AC, the vertical bisector De of AB intersects AC at point E, the perpendicular foot is D, and the vertical bisector of CE is just right When passing through point B, the perpendicular foot is f, then the degree of ∠ A is—— I'm in a hurry. I still have a lot of homework. There are only a few questions left Help, I'm too lazy to do it. You should draw it out. The answer is 36 degrees. How can you figure out what it is like, give it to you, and hope you can give it awesome.
- 2. As shown in the figure, it is known that in △ ABC, ab = AC, the vertical bisector De of AB intersects AC at point E, and the vertical bisector of CE just passes through point B and intersects AC at point F. the degree of ∠ A is calculated
- 3. As shown in the figure, it is known that in △ ABC, ab = AC, the vertical bisector De of AB intersects AC at point E, and the vertical bisector of CE just passes through point B and intersects AC at point F. the degree of ∠ A is calculated
- 4. As shown in the figure, in △ ABC, it is known that ∠ DBC = 60 ° AC > BC, and △ ABC ′, △ BCA ′, △ cab ′ are equilateral triangles outside △ ABC shape, and point D is on AC, and BC = DC (1) prove: △ C ′ BD ≌ Δ B ′ DC; (2) prove: △ AC ′ D ≌ Δ DB ′ a
- 5. As shown in the figure, in the triangle ABC, angle a is equal to 60 degrees, angle B is equal to 75 degrees, ab = 4cm, find C of BC
- 6. As shown in the figure, ad is the middle line of △ ABC, point E is on the extension line of BC, CE = AB, ∠ BAC = ∠ BCA, verification: AE = 2ad
- 7. As shown in the figure, in the triangle ABC, ab = BC, ad is the middle line on BC, extend BC to point E, make CE = BC, prove: AE = 2ad
- 8. As shown in the figure, in the triangle ABC, the angle BAC = 90 ° and BD is the bisector of the angle ABC. The extension line of BD is perpendicular to the line passing through point C and E, and the extension line of CE intersecting Ba is F. the proof is: BD = 2ce Please help me
- 9. In RT △ ABC, ∠ C is equal to 90 degree CD.CE If ∠ A is equal to 30 °, try to judge the shape of ∠ BCE Geometric steps, is an equilateral triangle
- 10. In RT △ ABC, CD is the bisector of BCE. BCE = 60 °∠ 3 = ∠ 4. Describe the relationship between CB and CE. What's your conjecture? And prove it Note that it's in the RT triangle
- 11. As shown in the figure, in △ ABC, ab = AC, points D and E are two points on the edge of BC, and ad = BD, AE = CE. If ∠ DAE = 60 °, calculate the degree of ∠ BAC
- 12. As shown in the figure, △ ABC is an equilateral triangle, ∠ DAE = 120 ° to prove: (1) △ abd ∽ ECA; (2) BC2 = DB · CE
- 13. As shown in the figure, point D is a point on the AC side of equilateral △ ABC, ∠ 1 = ∠ 2, BD = CE
- 14. As shown in the figure, point D is a point on the AC side of equilateral △ ABC, ∠ 1 = ∠ 2, BD = CE
- 15. Triangle ABC is an equilateral triangle, D, B, C, e are on a straight line, angle DAE = 120 degrees, known BD = 1, CE = 3. Find the side length of an equilateral triangle
- 16. As shown in the figure, △ ABC and △ DBC, the angle BAC = angle BDC = 90 °, O is the midpoint of BC (1) △ AOD is isosceles triangle? Why?
- 17. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 °, point m is on edge BC, am = BM
- 18. In RT △ ABC, ∠ B = 90 ° is connected to the midpoint m of AC side, and BM = cm = am is proved
- 19. In triangle ABC, BD and CE are the bisectors of angle B and angle C respectively. Given be = CD, it is proved that triangle ABC is isosceles triangle I'm sorry, but I can imagine everything
- 20. Two bisectors BD and CE of triangle ABC intersect at point O, angle a = 60 degrees, proving: CD + be = BC There is an answer to this question on the Internet, but I can't understand it because of my low IQ. Dashen explains why ∠ EOB = 1 / 2 (180 ° - 60 °) = 60 ° This is a certain method: take a point F on the edge of BC, make BF = be, link of ∵ BD is the angular bisector, BF = be, Bo is the common edge, ∴△BEO≌△BFO →∠EOB=∠FOB=∠COD ∵∠A=60° The angle bisector is BD and CE ∵∠EOB=1/2(180°-60°)=60° Then ∠ COF = 180 ° - ∠ FOB - ∠ cod = 60 ° And ∵ CE is the angular bisector and CO is the common edge, ∠ COF = ∠ cod = 60 ° (proved) ∴△COF≌△COD →CF=CD ∴BC=BF+CF=BE+CD Or there are other ways,