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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
It is proved that: ? quadrilateral ABCD is a parallelogram, ad ∫ BC, 4 = ∠ 5, ? the bisector BF of ABC, 3 = ∠ 5, ? AF = AB, ∵ ad ∥ BC, ? 1 = ∠ AEB, ? the bisector of BAC, ? 1 = ∠ 2, ? 2 = ∵ AEB, ? be = AB, M be = AB, ? AB
In the triangle ABC, D is the middle point of B and C, M is the upper point of AD. the extension lines of BM and cm intersect AC, AB and E, f respectively In the triangle ABC, D is the midpoint of B and C, M is the point above ad. the extension lines of BM and cm intersect AC, AB and E, f respectively Verification: EF parallel BC Junior high school mathematics
It is proved that the parallel lines of ad passing through B and C respectively intersect the extension lines of CF and be respectively at M and n
The quadrilateral mbcn is a parallelogram
Let cm and BN intersect at point O. from MB ∥ Ao ∥ CN, we get: of / FM = OA / BM, OE / en = OA / CN
And BM = CN
So: of = FM
So: Mn ‖ EF
And Mn ‖ BC
So: EF ‖ BC
In the triangle row ABC, BM and CN are the angular bisectors of the triangle ABC. AE is perpendicular to BM to e, AF is perpendicular to CN to F, and EF is parallel to BC
Prolonged AF crossed BC at P and prolonged am crossed BC at Q
Because BM is called bisector, angle ABM = CBM, be = be, angle AEB = qeb = 90 degrees
So the triangle ABM is all equal to MBq, and AE = Eq
Similarly, AF = FP
Fe is the median line of the triangle Apq
So EF is parallel to BC
It is known that in △ ABC, BD bisects ∠ ABC, ed ‖ BC, EF ‖ AC, Confirmation: be = CF
∵ED∥BC，EF∥AC，
The quadrilateral EFCD is a parallelogram,
∴ED=CF，
∵ BD bisection ∠ ABC, EBD = ∠ FBD,
Ed ‖ BC, EDB = ∠ FBD,
∴∠EBD=∠EDB，
∴EB=ED，
∴EB=CF．
It is known that: in △ ABC, the bisectors of ∠ B and ∠ C intersect at point D, and pass D as EF ‖ BC, AB at point E and AC at point F. verification: be + CF = EF
Proof: ∵ BD bisection ∠ ABC,
∴∠EBD=∠DBC，
∵EF∥BC，
∴∠EDB=∠DBC，
∴∠EDB=∠EBD，
∴DE=BE，
Similarly, CF = DF,
∴EF=DE+DF=BE+CF，
Be + CF = EF
In the triangle ABC, the vertical bisector of angle ACB = 90 BC crosses BC at point D, AB at point E F, on the extension line AF = ce of De, find that ACEF is a parallelogram
According to the question, CE = be = AE = AF
In △ ABC, the vertical bisector of ∠ ACB = 90 ° BC intersects BC at D, AB at point E, f at point de, and AF = CE to prove that the quadrilateral ACEF is a parallelogram
Because of the CD = BD, the △ CDE ≌ CDE ≌ (BDE) is the ≌ CD ≌ BD ≌ CE ≌ CD ≌ B ≌ B ≌ B ≌ B ≌ B ≌ C ≌ B ≌ B ≌ CD ≌ B ≌ CD ≌ B ≌ CD 878087808780≌
As shown in the figure, in △ ABC, ∠ ACB = 90, D, e are the midpoint of sides BC and ab respectively. Extend De to point F, so that AF = CE, it is proved that the quadrilateral ACEF is a parallelogram 2) When the size of ∠ B satisfies what condition, the quadrilateral ACEF is diamond? Please answer and prove your conclusion 3) Can quadrilateral ACEF be square? Please explain the reason
1) Because D and E are the midpoint of BC and ab respectively
So DF / / AC
So angle FEA = angle CAE
Because the angle ACB = 90 degrees, e is the midpoint of ab
So CE = AE, angle CAE = angle ace
Because AF = CE, CE = AE
So AF = AE
So angle FEA = angle F
So the angle FAE = the angle AEC
So fa / / EC
So the quadrilateral ACEF is a parallelogram
2) When the angle B = 30 degrees, the quadrilateral ACEF is rhombic
It is proved that because the angle ACB = 90 degrees, e is the midpoint of ab
So EC = EB
So the angle ECB = angle B = 30 degrees
So the angle AEC = angle ECB + angle B = 60 degrees
Because CE = AE
So the triangle AEC is an equilateral triangle, EC = AC
So the parallelogram ACEF is diamond again
3) quadrilateral ACEF cannot be square
The reason is simple: the angle FEC cannot be a right angle
As shown in the figure, in △ ABC, ∠ ACB = 90 ° e is the midpoint of DF, the crossing point E is de ⊥ BC, and AF = CE, it is proved that the quadrilateral ACEF is a parallelogram
Because ∠ ACB = 90 ° so AC ⊥ BC, and de ⊥ BC, so AC parallel and de e are the midpoint of DF, so AC parallel with EF and AF = CE can be verified by using the parallelogram rule
(1) prove that the quadrilateral ACEF is a parallelogram. (2) when the size of B satisfies what conditions, the quadrilateral abef is diamond? Please answer and prove your conclusion. (3) can the quadrilateral ACEF be a square? Why?
Because De is the vertical bisector of BC, the angle bed is equal to the angle CED, and the angle CED is equal to the angle ace
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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
- 4. Known: as shown in the figure, ab ∥ CD, and ab + CD = BC, M is the midpoint of AD Confirmation: BM ⊥ cm
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- 12. AB equals BC equals CD equals de equals EF equals FG equals GA. if angle DAE is a, calculate a value In triangle ade
- 13. 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
- 14. 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
- 15. 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
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- 17. 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
- 18. 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
- 19. As shown in the figure, in the triangle ABC, the vertical bisector L1 of the AB side In the triangle ABC, the vertical bisector L1 of AB side intersects BC at D, L2 of AC side intersects BC at e, Li and L2 intersect at point O. the perimeter of triangle ade is 6 and that of OBC of triangle is 16. Find the length of Ao
- 20. As shown in the figure, in △ ABC, ∠ ACB = 90 °, point E is the midpoint of AB, connecting CE, passing through point E as ed ⊥ BC at point D, taking a point F on the extension line of De to make AF = CE