In RT △ ABC, ∠ B = 90 ° is connected to the midpoint m of AC side, and BM = cm = am is proved

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• 1. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 °, point m is on edge BC, am = BM
• 2. 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?
• 3. 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
• 4. As shown in the figure, point D is a point on the AC side of equilateral △ ABC, ∠ 1 = ∠ 2, BD = CE
• 5. As shown in the figure, point D is a point on the AC side of equilateral △ ABC, ∠ 1 = ∠ 2, BD = CE
• 6. As shown in the figure, △ ABC is an equilateral triangle, ∠ DAE = 120 ° to prove: (1) △ abd ∽ ECA; (2) BC2 = DB · CE
• 7. 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
• 8. 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
• 9. 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.
• 10. 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
• 11. 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
• 12. 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,
• 13. In triangle ABC, bisectors be and CD of angle ABC and angle ACB intersect at point O. if angle a = 60 °, what is the relationship between BC, CE and BD
• 14. As shown in the figure, in the triangle ABC, the angle a = 60 degrees. The bisectors BD and CE of the triangle ABC intersect at point F. if be = 4 and CD = 2, find the length of BC
• 15. It is known that be is the bisector of the inner angle of triangle ABC and CE is the bisector of the outer angle of triangle ABC Add a reason after each step
• 16. It is known that in the triangle ABC, ∠ B = 1 / 2 ∠ a, CD is perpendicular to BC, CE is the middle line on the edge BD, and AC = 2 / 1ab is proved Using the properties of right triangle (1)
• 17. In the triangle ABC, if the bisector of the inner angle B and C intersects at point E and the bisector of the outer angle intersects at point F, and angle a is equal to 70 degrees, then angle BEC equals? Angle BFC equals?
• 18. As shown in the figure, in △ ABC, ab = AC, point E is on the extension line of Ca, and ED ⊥ BC is on D
• 19. As shown in the figure, in △ ABC, ∠ C = 90 °, CD ⊥ AB, CM bisects AB, CE bisects ∠ DCM, then the degree of ∠ ace is______ ．
• 20. As shown in the figure, in △ ABC, ∠ C = 90 °, ad bisects ∠ BAC, ab = 5, CD = 2, then the area of △ abd is () A. 2B. 5C. 10D. 20