As shown in the figure, AC bisects ∠ DAB, ab ＞ ad, CB = CD, CE ⊥ AB in E, (1) Results: ab = AD + 2EB; (2) If ad = 9, ab = 21, BC = 10, find the length of AC

As shown in the figure, AC bisects ∠ DAB, ab ＞ ad, CB = CD, CE ⊥ AB in E, (1) Results: ab = AD + 2EB; (2) If ad = 9, ab = 21, BC = 10, find the length of AC

(1) It is proved that the extension line ad is in F by crossing C as CF ⊥ AD and crossing ad,
∵ AC is the bisector of ∵ DAE, CE ⊥ AB, CF ⊥ AF,
∴CE=CF，
In RT △ CFD and RT △ CEB
CF＝CE
CD＝CB ，
∴Rt△CFD≌Rt△CEB（HL），
∴FD=EB，
In RT △ CFA and RT △ CEA
CF＝CE
AC＝AC ，
∴Rt△CFA≌Rt△CEA（HL），
∴AF=AE，
Then AB = AE + EB = AF + EB = AD + DF + EB = AD + 2EB;
（2）∵AD=9，AB=21，
AB = AD + 2EB from (1) and 9 + 2EB = 21 from (1),
EB = 6,
∴AE=AB-EB=21-6=15，
And ∵ BC = 10,
In RT △ CEB, according to Pythagorean theorem, the following results are obtained
CE=
BC2−BE2=8，
In RT △ ace, according to Pythagorean theorem, the following results are obtained
AC=
AE2+CE2=17．

As shown in the figure, ∠ DAB = 60 °, CD ⊥ ad, CB ⊥ AB, and ab = 2, CD = 1, find the length of AD and BC As the title

Make the extension line of BC and ad to point E
Because CB ⊥ AB, ∠ DAB = 60 ⊥
So AEB = 30 degrees
Because CD = 1, CD ⊥ ad
So de = radical 3 * CD, CE = 2 * CD
So de = radical 3, CE = 2
In the triangle Abe, ab = 2
AE = 2Ab = 2 * 2 = 4, be = 2 times root 3
AD=AE-DE
=4-radical 3
BC=BE-CE
=Double root 3-2

As shown in the figure, ∠ DAB = 90 ° CD ⊥ ad, CB ⊥ AB, and ab = 2, CD = 1, find the length of AD and BC

Make the extension line of BC and ad to point E
Because CB ⊥ AB, ∠ DAB = 60 ⊥
So AEB = 30 degrees
Because CD = 1, CD ⊥ ad
So de = radical 3 * CD, CE = 2 * CD
So de = radical 3, CE = 2
In the triangle Abe, ab = 2
AE = 2Ab = 2 * 2 = 4, be = 2 times root 3
AD=AE-DE
=4-radical 3
BC=BE-CE
=Double root 3-2

As shown in the figure, AC bisects ∠ DAB, ab ＞ ad, CB = CD, CE ⊥ AB in E, (1) Results: ab = AD + 2EB; (2) If ad = 9, ab = 21, BC = 10, find the length of AC

(1) It is proved that: the extended line ad, crossing C as CF ⊥ ad, leads to the extension line at F, ∵ AC is the bisector of ⊥ DAE, CE ⊥ AB, CF ⊥ AF, ᚉ CE = CF, in RT △ CFD and RT △ CEB, CF = CECD = CB, ≌ RT △ CEB (HL), ∵ FD = EB, and in RT △ CFA and RT △ CEA, CF = CEAC = AC

1. In the isosceles trapezoid ABCD, if AB parallel DC, ∠ B = 60, AC bisection ∠ DAB, DC = 3, then trapezoid perimeter 2. If the upper and lower bottoms of isosceles trapezoid are known to be 4 and 7 respectively, and one angle is 120, then the waist length is 3. In trapezoid ABCD, ad is parallel to BC, and the diagonal intersects with O, Bo = 3Do, and the area of △ AOD is 4, then the area of △ AOB is 4. In isosceles trapezoid ABCD, ad is parallel to BC, ad = 6, ab = 7, BC = 13, then ∠ a=

(1)  trapezoid ABCD is isosceles trapezoid  bottom angle lower  DAB = ∠ B = 60 ° and

As shown in the figure, in trapezoidal ABCD, ab ‖ CD, ad = DC, it is proved that AC is the bisector of ∠ DAB

Proof: ∵ ab ∥ CD,
∴∠CAB=∠DCA．
∵AD=DC，
∴∠DAC=∠DCA．
∴∠DAC=∠CAB，
That is, AC is the angular bisector of DAB

2. In trapezoidal ABCD, ab ‖ CD, AC bisection ∠ DAB, DC: ab = 1:1.5, In trapezoidal ABCD, ab ‖ CD, AC bisection ∠ DAB, DC: ab = 1:1.5, Then ad: BC =

One hundred and twenty-three

In the isosceles trapezoid ABCD, AB / / DCAD = DC = BC = 10 angle DAB = 60 degrees is used to calculate the trapezoid area

Make de ⊥ AB at point E and DF ⊥ AB at point F
Then EF = CD = 10
∵∠B=60°
∴AE=5
According to Pythagorean theorem, de = 5 √ 3 can be obtained
BF = 5 can be obtained by exerting force
∴AB=5+5+10=20
ν s trapezoid ABCD = 1 / 2 (10 + 20) * 5 √ 3
=75√3

In the isosceles trapezoid ABCD, AB is parallel to DC, AC bisects ∠ DAB, ∠ DAB = 60 ° if trapezoid perimeter is 8, what is ad?

Because ∠ DAB = 60ac bisects ∠ DAB, so ∠ DAC = ∠ BAC = 30, so the triangle ABC is a right triangle, where ∠ B = 60, so let BC = ad = a, so AB = 2A AC = √ 3a, because ∠ DAC = ∠ dcb-acb = 30 = DAC, so ad = DC = a, so perimeter = AB + BC + DC + ad = 2A + A + A + a = 5A = 8, so a = 8 / 5, so ad = 8 / 5

As shown in the figure, angle DAB + angle CDA = 180 degrees, angle ABC = angle 1, is straight line AB parallel to CD? What about straight line AD and BC? Why?

Parallel because angle DAB + angle CDA = 180 degrees, DC is parallel to ab
Because the angle ABC = angle 1, ad is parallel to BC (according to the property of parallelism)