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As shown in the figure, △ ABC, ab = AC, points m and N are on the straight line where BC is located, and am = an Excuse me: BM = CN? Please state the reasons
BM=CN．
Reason: ∵ AB = AC,
∴∠B=∠C，
And ∵ am = an,
∴∠AMN=∠ANM，
∴∠AMB=∠ANC，
∴△ABM≌△ACN，
∴BM=CN．
As shown in the figure, △ ABC, ab = AC, points m and N are on the straight line where BC is located, and am = an Excuse me: BM = CN? Please state the reasons
BM=CN．
Reason: ∵ AB = AC,
∴∠B=∠C，
And ∵ am = an,
∴∠AMN=∠ANM，
∴∠AMB=∠ANC，
∴△ABM≌△ACN，
∴BM=CN．
As shown in the figure, △ ABC, ab = AC, points m and N are on the straight line where BC is located, and am = an Excuse me: BM = CN? Please state the reasons
BM=CN．
Reason: ∵ AB = AC,
∴∠B=∠C，
And ∵ am = an,
∴∠AMN=∠ANM，
∴∠AMB=∠ANC，
∴△ABM≌△ACN，
∴BM=CN．
As shown in the figure, in the triangle ABC, ab = AC, points m, N, respectively in BC line and am = an. Ask, BM = CN, do? (not congruent)
Draw the edge of ad vertical BC triangle ABC
There is also triangle amn DM = DN, so DB = DC, empathy = CN
As shown in the figure, triangle ABC is an equilateral triangle, points m and N are on BC and AC respectively, and BM = CN, am and N intersect at point Q. find the degree of angle aqn
In △ ABM and △ BCN: ab = BC, ∠ ABC = ∠ ACB = 60 °, in △ ABM and △ BCN: ab = BC, ∠ ABC = ∠ ACB, BM = cn ABM is all equal to △ BCN (SAS) ? from the problem: ? AMB = ∠ AMC = 90 °, in RT △ ABM and RT △ ACM: ab = AC, am = am RT △ ABM are all equal to RT
In trapezoid ABCD, AD / / BC, ab = CD, e, F, G, h are the midpoints of each side. It is proved that the quadrilateral efgh is a diamond In trapezoid ABCD, AD / / BC, ab = CD, e, F, G, h are the midpoint of each side (1) Verification: quadrilateral efgh is rhomboid (2) If the quadrilateral ABCD is a rectangle, and E, F, G, h are still the midpoint of each side, is the quadrilateral efgh still a diamond? Why
In trapezoid ABCD, AD / / BC, ab = CD, e, F, G, h are the midpoint of each side. (1) it is proved that the quadrilateral efgh is a diamond. Connecting AC, BD, E on AB, f on BC, G on CD, h on ad, because e, F, G, h are the midpoint of each side, Hg = 1 / 2Ac, EF = 1 / 2Ac, so Hg = EF, Hg ‖ EF is the same as eh = GF, eh ‖ GF
It is known that: in trapezoid ABCD, ad is parallel to BC, ab = CD, and mnef is the midpoint of edge ad BC, AB DC, respectively
It is known that MF is the median line of △ DAC, so MF is parallel to AC and equal to 1 / 2Ac, en is the median line of △ BAC, so en is parallel and equal to 1 / 2Ac, so MF is parallel to en and equal to 1 / 2Ac. In trapezoid ABCD, ad is parallel to BC, ab = CD, so trapezoid is isosceles trapezoid, then ∠ B = ∠ C, and ab =
M. The results show that BM = DN, me is vertical BD, NF is vertical BD, vertical foot is e, F, Mn and EF are equal
Step 1: it can be proved that △ den ≌ △ BFM [AAL] - → ne = MF
The second step: prove that △ ENO ≌ △ FMO (EF, MN intersect O) [AAL] - → NO = MO, EO=FO,
As shown in the figure. In ▱ ABCD, AE ⊥ BC, CF ⊥ ad, DN = BM
It is proved that because ABCD is a parallelogram, so ad ≌ BC, and ad = BC, ab ∥ CD, and ab = CD, ⊥ BC, CF ⊥ ad, so aecf is rectangular, so AE = cf. so RT ≌ Abe ≌ RT △ CDF (HL, or AAS), be = DF
It is known that AC is a diagonal of the parallelogram ABCD, BM ⊥ AC, DN ⊥ AC, and the perpendicular feet are m and N respectively, Verification: quadrilateral bmdn is parallelogram
Proof: BM ⊥ AC, DN ⊥ AC,
∴∠DNA=∠BMC=90°，
∴DN∥BM，
∵ quadrilateral ABCD is a parallelogram,
∴AD∥BC，AD=BC，
∴∠DAN=∠BCM，
∴△ADN≌△CBM，
∴DN=BM，
The quadrilateral bmdn is a parallelogram
RELATED INFORMATIONS
- 1. In △ MNB, BN = 6, points a, C, D are on MB, Nb, Mn respectively, and the quadrilateral ABCD is a parallelogram, and ∠ NDC = ∠ MDA, then the circumference of quadrilateral ABCD is () A. 24 B. 18 C. 16 D. 12
- 2. Known: as shown in the figure, ab ∥ CD, and ab + CD = BC, M is the midpoint of AD Confirmation: BM ⊥ cm
- 3. She is an English girl
- 4. Lili is___ ChinEse and Jim is___ How to choose American boy A. /; a B.A. / c.an; a D.A; an
- 5. you do not come from England,do you?----____ .I come from America.How do you know that?
- 6. I'm an American boy
- 7. I come from China. Or I'm come from China?
- 8. Plane PAC ⊥ plane ABC, PA = Pb = PC, verify ab ⊥ BC
- 9. It is known that P is a point in the plane of triangle ABC. If vector CB = γ vector PA + vector Pb, γ belongs to R, then point P must be in the plane A. Inside the triangle ABC B. On the line where the AC side is located C. On the line where AB side is located D. On the line where the BC side is located Why?
- 10. As shown in the figure, BC is the diameter of ⊙ o, P is a point on ⊙ o, a is the midpoint of arc BC, ad ⊥ BC, perpendicular foot is D, Pb intersects AD and AC respectively at points e, F, AE and be? Why? It stops after 20 minutes
- 11. As shown in the figure, it is known that m and N are two points on the edge BC of △ ABC, and BM = Mn = NC is satisfied. A straight line parallel to AC intersects the extension line of AB, am and an at points D, e and f respectively. It is proved that EF = 3DE
- 12. As shown in the figure, in the parallelogram ABCD, the bisector of ∠ bad intersects with edge BC at point E, and bisector of ∠ ABC intersects with edge ad at point F. please prove that the abef of quadrilateral is diamond
- 13. In the triangle ABC, D is the midpoint of BC and DM is perpendicular to DN. If BM + CN = DM + DN, verify Ad = 1 / 4 (AB + AC)
- 14. AB equals BC equals CD equals de equals EF equals FG equals GA. if angle DAE is a, calculate a value In triangle ade
- 15. In the graph + equilateral triangle ABC, if D, e and F are on BC, AC and ab respectively, and CD = BF, BD = CE, then what degree is the angle EDF
- 16. It is known that ad is an angular bisector of △ ABC, where de ‖ AC intersects AB at point E and DF ‖ AB intersects AC at point F Verification: quadrilateral AEDF is rhombic
- 17. As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = 4, AC = 4, now translate △ ABC along CB direction to △ a ′ B ′ C ', if the translation distance is 3 (1) Calculate the area of the overlapped part of △ ABC and △ a ′ B ′ C '; (2) If the translation distance is x (0 ≤ x ≤ 4), find the area y of the overlapping part of △ ABC and △ a ′ B ′ C ', then what is the relationship between Y and X
- 18. It is known that in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of AC, AE ⊥ BD is at e, and extended AE is crossed with BC at F. it is proved that ∠ ADB = ∠ CDF
- 19. As shown in the figure, the turning point of a highway is an arc. The point O is the center of the arc, AB is equal to 120m, C is the point on AB, O is the point on AB, the perpendicular foot is D, and CD is equal to 20m. Calculate the radius of the curve
- 20. It is known that in the square ABCD, ∠ man = 45 ° and ∠ man rotate clockwise around point A. both sides of it intersect CB, DC (or their extension) at point m and N, ah ⊥ Mn at point H (1) As shown in Fig. 1, when ∠ man point a is rotated to BM = DN, please write down the quantitative relationship between ah and ab directly______ ; (2) As shown in Fig. 2, when ∠ man rotates around point a to BM ≠ DN, does the quantitative relationship between ah and AB in (1) still hold? If not, please write down the reason, if yes, please prove it; (3) As shown in Fig. 3, given ∠ man = 45 °, ah ⊥ Mn at point h, and MH = 2, NH = 3, find the length of ah