![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
AB = AC = ad, ∠ DAC = 3 ∠ cab Angle BDC = 1 / 2 angle cab, angle DBC = 1 / 2 angle DAC How did you get this step?
Because AB = AC = ad
So B, C and D are on the circle with a as the center and ab as the radius
Because the circle angle corresponding to the same arc is half of the corresponding center angle
Therefore, BDC = 1 / 2 cab and DBC = 1 / 2 DAC
Because ∠ DAC = 3 ∠ cab
Therefore, DBC = 3 BDC
Therefore, BDC = 1 / 3 DBC
As shown in the figure, in the quadrilateral ABCD, ab = AC = ad, (1) if ∠ DAC = 2 ∠ BAC, then ∠ DBC = 3 ∠ BDC, explain the reason
Because AB = AC = ad, ∠ ABC = ∠ ACB, ∠ ACD = ∠ ADC
Because ∠ DAC = 2 ∠ BAC, if ∠ BAC = x, then ∠ DAC = 2x
Therefore, ABC = ACB = 90-x, ACD = ADC = 90 - (1 / 2) x, abd = ADB = 90 - (3 / 2) X
Then ∠ DBC = 90-x-90 + (3 / 2) x = (1 / 2) x, ∠ BDC = 90 - (1 / 2) x-90 + (3 / 2) x = X
Therefore, DBC = 2 BDC
In the quadrilateral ABCD, ab = AC = ad, ∠ DAC = 2 ∠ BAC,
To prove this problem, we mainly use the angle transformation, because ∠ DAC = 2 ∠ BAC, so we can transform the angle into n ∠ BAC. Prove: because AB = AC = ad, in triangle abd, ∠ ADB = abd = (180 - ∠ bad) / 2 = (180 - ∠ BAC - ∠ DAC) / 2 = (180-3 ∠ BAC) / 2 & nbsp; & nbsp; in triangle ACD, ∠ ADC = ∠ ACD = (180 - ∠ DAC) / 2 = (180-2 ∠ BAC) / 2 & nbsp; & nbsp; in triangle ABC, In triangle BCD, ∠ ABC = ∠ ACB = (180 - ∠ BAC) / 2 & nbsp; & nbsp; in triangle BCD, ∠ BDC + ∠ DBC = 180 - ∠ ACB - ∠ ACD = = 3 ∠ BAC / 2, and because ∠ abd + ∠ DBC = ∠ ACB & nbsp; so, ∠ ABC = ∠ ACB - ∠ abd = (180 - ∠ BAC) / 2 - (180-3 ∠ BAC) / 2 = ∠ BAC & nbsp; so, ∠ BDC = ∠ BAC / 2 & nbsp; & nbsp; that is, ∠ DBC = 2 ∠ BDC, it is proved
As shown in the figure, in the quadrilateral ABCD, the diagonal AC and BD intersect at point E, and
In the square ABCD, E.F is the point on the side AB, BC of the square, AE + CF = EF
I know that we should use the bisector to prove that it's the less conditional graph. Please draw it. Thank you
I haven't learned Tan α = x / A, Tan β = Y / A, set Tan (α + β) = (Tan α + Tan β) / (1-tan α · Tan β) and get some answers
Connect de and DF, rotate the triangle DAE clockwise 90 degrees with D as the rotation center, e falls on the extension line h of BC, so de = DH, because AE + CF = EF, AE = ch, so EF = CF + CH, that is to say
EF = FH, de = DH, EF = FH, DF = DF, triangle def, DHF, congruent, angle EDF = angle HDF, you should know after it
As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, e and F are the midpoint of AB and AC respectively, BD and EF intersect g. prove GF = (BC-AD) / 2
It is proved that AE = EB, AF = FC
∴EF∥=BC/2
AD∥BC ∴EF∥AD
Ψ BG = Gd (a line passing through the midpoint of one side of a triangle and parallel to the other side must bisect the third side)
∴EG∥=AD/2
GF = ef-eg
∴GF=(BC/2)-(AD/2)=(BC-AD)/2
Known: as shown in the figure, in the trapezoidal ABCD, AD / / BC / / EF, diagonal dB and AC intersect at point O, and EF intersect at point h, g respectively
Certification:
∵AD//BC//EF
Ψ AE / be = EF / CF
That is, (AE + be) / be = (EF + CF) / CF
∴AB/BE=DC/CF
∵AD//EF
∴⊿BEH ∽⊿BAD,⊿CFG∽⊿CDA
∴AB/BE=AD/EH,DC/FC=AD/FG
∴AD/EH=AD/FG
∴EH=FG
In space quadrilateral ABCD, e, F, G and H are the midpoint of AB, BC, CD and Da respectively, and AC = BD. it is proved that efgh is a plane graph
EF is parallel to AC and equal to half of AC, GH is parallel to AC and equal to half of AC, so EF is parallel to GH. Similarly, FG is parallel to he, so efgh is in the same plane
As shown in the figure, e, F, G and H are the midpoint of AB, BC, CD and DA in space quadrilateral ABCD. Verification: 1) eh / / FG, eh = FG; 2) AC / / plane efgh
If BD is connected, EH is the median line because EH is the midpoint, so eh / / BD, EF = 1 / 2bd. Similarly, GF / / BD, GF = 1 / 2bd, so Ed / / GF and ED = GF. If EF is the median line, then EF / / AC, EF belongs to face efgh, AC does not belong to face efgh, so AC / / face efgh
In the square ABCD, e is any point on the diagonal BD. if the circumference of the square ABCD is
16, find the perimeter of EFCG
EG=DG EF=CG
Eg + EF = square side length a
ABCD circumference = 4A = 16
a=4
So EFCG perimeter = 2A = 8
RELATED INFORMATIONS
- 1. 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,]
- 2. 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
- 3. It is known that as shown in the figure, ad and BC intersect at point O, ∠ cab = ∠ DBA, AC = BD
- 4. 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?
- 5. In the isosceles triangle ABCD, ab = AC, ∠ a = 36 ° and BD is the square of ∠ ABC, then calculate AD / AC
- 6. As shown in the figure, in the square ABCD, point P is a point on the diagonal AC, PE ⊥ AB, PF ⊥ BC, and the perpendicular feet are points E and f respectively. If the perimeter of the square ABCD is 8cm, then the perimeter of the quadrilateral EBFP is______ cm．
- 7. It is known that in trapezoidal ABCD, ad is parallel to BC, AB is parallel to de, AF is parallel to DC, e and F are on edge BC As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, ab ‖ De, AF ‖ DC, e and F are on the side BC, and the quadrilateral aefd is a parallelogram (1) What is the equivalent relationship between AD and BC (2) When AB = DC, it is proved that the quadrilateral aefd is a rectangle I want to prove that the process is not an equation
- 8. As shown in the figure, in trapezoidal ABCD, ad is parallel to BC, AF is vertical, BC and points F, m are the midpoint of CD, angle B is 45 degrees, AF is 4, EF is 7 Find ABCD area
- 9. In the trapezoidal ABCD, ab = CD, e is the midpoint of AD
- 10. 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
- 11. Isosceles trapezoid ABCD, ab = CD = 5, ad = 4, BC = 10, e, f on AB, BC, EF bisects the circumference of trapezoid, BF = x, and uses X to express the area of △ bef AB DC is waist
- 12. As shown in the figure, in square ABCD, point m is a point on diagonal BD. through point m, make me parallel to CD, intersect BC at point E, and make MF parallel to BC at point F. prove am = EF As shown in the figure, in the square ABCD, point m is a point on the diagonal BD, passing through point m to make a point Me parallel CD intersects BC at point E, MF parallel BC at point F. prove am = EF
- 13. 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
- 14. 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
- 15. As shown in the figure, in ladder ABCD, ab ‖ DC and E are the midpoint of BC, and the extension lines of AE and DC intersect at point F to connect AC and BF. What quadrilateral is abfc? Please give reasons
- 16. As shown in the figure, the quadrilateral ABCD is a square, AB is an equilateral triangle with an edge to the square, Abe, CE and DB intersect at point F, then ∠ AFD=______ Degree
- 17. Given that the length of a diagonal line of a diamond is 4cm and the perimeter is 16cm, what are the degrees of the four inner angles of the diamond?
- 18. If the perimeter of the diamond is 8cm and the height is 1cm, the angle of the diamond is smaller______ ．
- 19. As shown in the figure, the parallelogram ABCD, e is the midpoint of AD, AC and be intersect at the point F, s △ EFC = 1, then s parallelogram ABCD=
- 20. In rectangular trapezoid ABCD, ad is parallel to BC, ∠ a = 90 degrees, ab = ad = 1 / 2BC, and m and N are the midpoint of AD and DC respectively, connecting BM, Mn and BN. Excuse me: triangle What is the triangle of BmN