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As shown in the figure, is the line AB parallel to CD? And the line AD and BC? Why? Because of what, so what
∵∠DAB+∠CDA=180°
/ / AB ∥ CD (complementary to the inner corner of the same side and parallel to the two lines)
∵∠ABC=∠1
/ / ad ∥ BC (equal position angle, two parallel lines)
As shown in the figure, in the parallelogram ABCD, bisectors of ∠ DAB and ∠ CDA intersect with E, and bisectors of ∠ ABC and ∠ BCD intersect with F. it is proved that EF = ab-bc
The cross between BC and G was prolonged by AE, and AB was crossed with h by CF
In the parallelogram ABCD
Angle a = angle c, AE bisection angle a CF bisection angle c
Angle BCF = angle BAE
AB parallel CD
Angle DGE = angle BAE
Angle DGE = angle BCF
Angle B = angle D de bisection angle D BF bisection angle B
Angle EDG = angle FBC
Ed = bf because of parallelism
BCF≌EDG
DG=BC GE=CF
Because Ag ‖ CF
EF‖=CG
CG=DC-DG
EF=AB-BC
Factorization (BCD + CDA + DAB + ABC) ^ 2 - (BC AD) (CD AB) (BD AC)
(bcd+cda+dab+abc)²-(bc-ad)(cd-ab)(bd-ac)
=[bd(c+a)+ac(b+d)]²-b²d²(a²+c²)+abcd(b²+d²)+abcd(a²+c²)-a²c²(b²+d²)
=b²d²(c+a)²+a²c²(b+d)²+2abcd(a+c)(b+d)-b²d²(a²+c²)+abcd(b²+d²)+abcd(a²+c²)-a²c²(b²+d²)
=2acb²d²+2bda²c²+abcd(a²+b²+c²+d²+2ab+2ad+2bc+2cd)
=abcd(a²+b²+c²+d²+2ab+2ad+2bc+2cd+2bd+2ac)
=abcd[(a+c)²+(b+d)²+2(a+c)(b+d)]
=abcd(a+c+b+d)²
AB is parallel to CD and BC is parallel to ad, indicating that angle B is equal to angle D
prove:
∵AB∥CD
Ψ D + ∠ a = 180 ° (two straight lines are parallel, and the inner angles of the same side are complementary)
∵AD∥BC
Ψ B + ∠ a = 180 ° (two straight lines are parallel, and the inner angles of the same side are complementary)
∴∠D+∠A＝∠B+∠A
∴∠D＝∠B
AB is parallel to CD, ad is parallel to BC. Two methods are used to show that angle DAB is equal to angle BCD Two ways
Method (1) because AB parallels CD and ad parallels BC, so quadrilateral ABCD is parallelogram, so angle DAB = angle BCD
Method (2), because AB is parallel to CD, the angle DAB + angle ABC = 180 degrees,
Because ad is parallel to BC, angle BCD + angle ABC = 180 degrees,
So angle DAB = angle BCD
Angle 1 equals angle 2. Angle 3 equals angle 4, indicating that ad is parallel to be
Man, how can I answer without a picture?
In trapezoidal ABCD, AB is parallel to CD, angle a is equal to 90 degrees, AB is equal to 2, BC is equal to 3, CD is equal to 1, e is the midpoint of AD, and CE is perpendicular to be It seems that I haven't learned what you two said. I don't know what the symbol is. Use Pythagorean theorem So ad = FC = 2, root number 2 So AE = ed = radical 2 AB^2+AE^2=EB^2=6 ED^2+DC^2=EC^2=3 EB^2+EC^2=9=BC^2 Prove CE vertical be
CF ⊥ AB in F ∵ a = 90 ° ab ? a = 90 ° ab ∵ CD d = 90 ° ? BF = ab-af = ab-cd = 2-1 = 1 ? CF = √ (BC ⊥ BF ⊥ CF = ad = 2 √ 2, AE = de = √ 2 ? CE = √ 3, be = √ (AB ∧ AE ᙨ 6 BC B ⊥ B ⊥ CD ? d = 90 °
As shown in the figure, in right angle trapezoid ABCD, ad is parallel to BC, ∠ a = 90 °, BD = CD, point E is the point above AB, and ∠ AED = ∠ Dec (1) Results: BC = 2ad; (2) Explore: the quantitative relationship between ce-be and AE, and prove
(1) If the crossing point D is DF ⊥ BC, then ad = BF
∵BD=CD
∴BF=CF=AD
BC = 2ad
(2) Pass through point D as DP ⊥ CE at point P, and take a point Q on CE so that PQ = PE
Then there is △ ade ≌ △ PDE ≌ △ PDQ
∴EQ=2AE,DE=DQ,∠AED=∠QDP①
And ∵ ad = DP, BD = CD, ∵ a = ∵ DPC
∴△ABD ≌ △PDC
∴∠ADB=∠PDC②
From ① and ② we know that ∠ EDB = ∠ QDC
DQ = De, DC = BD
∴△EDB ≌ △QDC
∴BE=CQ
To sum up, Ce - be = 2ae
As shown in the figure, ab ∥ DC, e are the midpoint of BC, AED = 90, try to explain AB + CD = ad
EF / / AB by e
Then f is the mid point of AD
So EF is half of the hypotenuse of a right triangle
EF = 1/2AD
meanwhile
EF is the midline of trapezoidal waist = 1 / 2 (AB + CD)
therefore
AB+CD=AD
In the trapezoid ABCD, AB is parallel to CD, e is the midpoint of BC, and the angle AED is equal to 90 degrees
If the extension line of de intersecting AB is at F, then △ DCE ≌ △ FBE, ᚔ DC = FB, de = Fe
It can be proved that △ ade ≌ △ AFE, ᚔ ad = AF
RELATED INFORMATIONS
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