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As shown in Figure 1, in quadrilateral ABCD, ad ∥ BC, de bisects ∥ ADB, ∥ BDC = ∥ BCD, (1) (2) if the bisector of abd and the extension of CD intersect at f (Fig. 2), and F = 55 °, find ABC
(1) In the △ BCD, the = 90 ° - F= 90 ° - 55 ° = 35 ° according to the external angle property of triangle, ∠ OBD + ∠ ODB = ∠ DOF = 35 °, ∵ de bisecting ∠ ADB, BF bisecting ∠ abd, ∵ abd + ∠ ADB = 2 (∠ OBD + ∠ ODBC) = 2 × 35 ° = 70 ° in △ abd, ∵ a = 180 ° - (∠ abd + ∠ ADB) = 180 ° - 70 ° = 110 °, ∵ ad ‖ BC, ∵ ABC = 180 ° - a = 180 ° - 110 ° = 70 °
As shown in the figure, in the quadrilateral ABCD, AD / / BC, de bisects ∠ ADB, ∠ BDC = ∠ BCD, and proves ∠ 1 + 2 = 90 degree
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AD∥BC,
∠ADC+∠BCD=180°,
∵ de bisection ∠ ADB,
∠BDC=∠BCD,
∴∠ADE=∠EDB,
∠BDC=∠BCD,
∵∠ADC+∠BCD=180°,
∴∠EDB+∠BDC=90°,
∠1+∠2=90°．
As shown in the figure, in the quadrilateral ABCD, ad ∥ BC, de bisects ∥ ADB, ∥ BDC = ∥ BCD
24. As shown in the figure, in the quadrilateral ABCD, ad ∥ BC, de bisection ∥ ADB, ∥ BDC = ∥ BCD,
(1) Verification: 1 + 2 = 90 degree
(2) If the bisector of ∠ abd intersects the extension of CD at F, and ∠ f = 55 °, find ∠ ABC
(1) It is proved that: ad ∥ BC, ∥ ADC + ∥ BCD = 180, ∥ de bisection ∥ ADB, ∥ BDC = ∥ BCD, ∥ ade = ∥ EDB, ∥ BDC = ∥ BCD, ∥ ADC + ∥ BCD = 180 °, ∥ EDB + ∥ BDC = 90 °, ∥ 1 + ∥ 2 = 90 ° (2) ∥ FBD + ∥ BDE = 90 ° - ∥ f = 35 °, ∥ de bisection ∥ ADB, BF bisection ∥ abd, ∥
In quadrilateral ABCD, AD / / BC, ad is not parallel BC, ∠ ACB = ∠ DBC, is quadrilateral ABCD isosceles trapezoid? Thank you,
It is proved that: ∵ cab = ≌ DBA, AC = BD, ab = Ba, ≌ ACB ≌ BDA. ≌ ad = BC, ≌ ABC = ≌ bad. If de ∥ BC intersects AB with E, then ∵ DEA = ∥ CBA, ∩ DAE = ∥ DEA. ≌ ad = ed. ≌ ed = BC, ≌ AED = ≌ ABC. ≌ ab ∥ CD is an isosceles trapezoid
Cut AE = BF = CG = DH on each side of square ABCD, connect AF, BG, CH, De, intersect n, P, Q, m in turn, prove that the quadrilateral mnpq is a square
Because it is a square, ab = BC = CD = ad and AE = BF = CG = DH, EB = FC = GD = ha, angles a, B, C, D are all right angles, so AEH, EBF, CFG, DHG are all equal to ef = FG = GH = he, so the quadrilateral mnpq is diamond and AEH = EFB, EFB + AEH = 90hef = 90, so the quadrilateral mnpq is square
Cut AE = BF = CG = DH on each side of square ABCD, connect AF, BG, CH and De, and intersect n, Q, P and D in turn
emergency
Cut AE = BF = CG = DH on each side of square ABCD, and connect e, F, G and h in turn. Ask: is efgh a square? Please explain the reason. It is a square proof: ∵ AE = BF = CG = DH ≌ ah = DG = CF = be and ∵ a = ∩ d = ∩ B = ∩ C ≌ AHG ≌ DGH ≌ CFG ≌ bef ≌ Hg = GF = EF = he
Given that the diagonal of square ABCD intersects at O, OE is perpendicular to of, AE = 4, CF = 3, find EF
It is known that in the parallelogram ABCD as shown in the figure, points E and F are on BC and DC respectively, and AE = AF, DG vertical, AE, BH vertical, AF, G and H are perpendicular feet
You didn't say which side E and F are on. First, I think that e is on AD and F is on CD, which doesn't affect the calculation results. Because OE is perpendicular to ED and DF is perpendicular, in quadrilateral eofd, angle deo + angle ofd = 180 degrees, and because angle deo + angle AEO = 180 degrees, "angle ofd = angle AEO" is square ABCD, so "angle oad =..."
In square ABCD, diagonal lines AC and BD intersect at point O, e and F are points on edge AB and BC respectively. If AE = 4cm, CF = 3cm, and OE is perpendicular to of, then EF=——
5
In square ABCD, O is the intersection of diagonal lines AC and BD, passing through o as OE ⊥ of, respectively intersecting AB and BC at e and F, AE = 12 and CF = 5, then the value of EF is
Let oe1 and of1 be vertical lines on AB and BC
Off1 and oee1 congruence
AE=AE1 + E1E
CF= CF1 - FF1
AE1=CF1
E1E=FF1
AE1 = (AE+CF)/2 = 8.5
FF1 = 3.5
OF = OE = sqrt(8.5^2 + 3.5^2)
EF = sqrt(2)*OF = 13
In the isosceles trapezoid ABCD, ab ∥ DC, ad = BC = 5, DC = 7, ab = 13, point P starts from point a and moves along ad → DC to terminal C at the speed of 2 units per second, while point Q starts from point B and moves along Ba to terminal a at the speed of 1 unit per second, and the movement time is set as T seconds
(1) When what is the value of T, the quadrilateral pqbc is a parallelogram?
(2) If point P is on DC, what is the value of T? In the whole movement process, the triangle with points c, P and Q as vertices is a right triangle?
If pqbc is a parallelogram, QB = CP,
Namely: 1 * t = DC + AD-2 * t
It can be concluded that: T = 4 (seconds)
If a triangle with vertices C, P and Q is a right triangle, then QP = 4, BQ = 6,
It can be concluded that: T = 6 / 1 = 6 (seconds)
RELATED INFORMATIONS
- 1. It is known that as shown in the figure, ad and BC intersect at point O, ∠ cab = ∠ DBA, AC = BD
- 2. The quadrilateral ABCD is the inner boundary quadrilateral of the circle, the diagonal AC and BD intersect at e, the extended DA and CB intersect at F, the angle CAD = 60 degrees, DC = De, it is proved that a is the outer center of the triangle bef How does angle F + angle AEB equal to 60 degrees?
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- 7. In the trapezoidal ABCD, ab = CD, e is the midpoint of AD
- 8. As shown in the figure, in trapezoidal ABCD, ∠ d = 90 ° and M is the midpoint of ab. if cm = 6.5 and BC + CD + Da = 17, the area of trapezoidal ABCD is () A. 20B. 30C. 40D. 50
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- 11. As shown in the figure, in the quadrilateral ABCD, the diagonal AC and BD intersect at point E, and ∠ cab = ∠ CBD. Given AB = 4, AC = 6, BC = 5, de = 5.5, find the length of de. & nbsp; & nbsp; [please solve this problem from the Perspective of similar triangles,]
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- 15. As shown in the figure, m and N are the midpoint of AD and DC on both sides of the square ABCD, cm and BN intersect at P
- 16. As shown in the figure, in the quadrilateral ABCD, ab = CD, points E and F are the midpoint of AD and BC respectively, GH ⊥ EF intersects point P
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