![](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>)
It is known that, as shown in the figure, in right angle trapezoid ABCD, ad is parallel to BC, ∠ ABC = 90 °, de ⊥ AC is at point F and BC is intersected at point G, which is called ab Verification: BG = FG No picture. Sorry
It is proved that if ad is parallel to BC, the angle DAC = angle ACB, angle ABC = angle DAE = 90 ° so angle e + angle ade = 90 ° de ⊥ AC, angle DAC + angle ade = 90 ° so angle DAC = angle e = angle ACB, angle ABC = angle AFE = 90 ° AE = AC, so triangle AFE is equal to triangle ABC, so AB = AF then be = FC can get triangle
In the right angle trapezoid ABCD, ad ‖ BC ∠ ABC = 90 ° AB = BC e is a point on the edge of AB and AE = ad link de intersects diagonal AC at point H ① Confirmation: ah ⊥ de ② If ∠ BEC = 75 °, it is proved that △ CDE is equilateral △
prove
①∵∠ABC=90° AB=BC
ν Δ ABC is an isosceles right triangle,
∴∠CAB=45°
∵ AE = ad ᙽ ead is an isosceles right triangle
∴∠AED=45°
cab = 45 ° ahe is also an isosceles right angle △
∴∠AEH=90°,∴AH⊥DE
②∠BEC=∠BAC+∠ACE=75°
∵∠BAC=45°,∴∠ACE=30°
﹤ CEH = 60 ° in right angle △ EHC
From AE = ad, ∠ EAC = ∠ DAC = 45 ° and AC = AC
We can get △ AEC ≌ △ ADC, ᚔ CE = CD
Ψ Δ CED is isosceles △, ∵ CEH = 60
There is an isosceles with an angle of 60 degrees, which is an isosceles
The △ CDE is an equilateral △
As shown in the figure, in right angle trapezoid ABCD, ad ∥ BC, ∠ ABC = 90 °, de ⊥ AC at point F, intersection BC at point G, extension line crossing AB at point E, and AE = AC (1) Results: BG = FG; (2) If ad = DC = 2, find the length of ab
(1) Proof: connect AG,
∵ ABC = 90 ° de ⊥ AC at point F,
∴∠ABC=∠AFE．
In △ ABC and △ AFE,
∠ABC＝∠AFE
∠EAF＝∠CAB
AC＝AE
∴△ABC≌△AFE（AAS），
∴AB=AF．
In RT △ ABG and RT △ AFG,
AG＝AG
AB＝AF
∴Rt△ABG≌Rt△AFG（HL）．
∴BG=FG；
（2）∵AD=DC，DF⊥AC，
ν f is the midpoint of AC,
∵AC=AE，
∴AF=1
2AC=1
2AE．
∴∠E=30°．
∵∠EAD=90°，
∴∠ADE=60°，
∴∠FAD=∠E=30°，
∴AF=
3．
∴AB=AF=
3．
As shown in the figure, in the right angle trapezoid ABCD, ∠ ABC = 90 °, ad ‖ BC, ad = 4, ab = 5, BC = 6, point P is a moving point on ab. when the sum of PC + PD is the smallest, the length of Pb is () A. 1 B. 2 C. 2、5 D. 3
If Da is extended to d ', then D and d' are symmetric about AB, connect CD ', and intersect with ab at point P,
According to "the shortest line segment between two points", the sum of PC+PD is minimum
Because of ad ′∥ BC, △ APD ∽ BPC
Let Pb = x, then AP = 5-x
So AP
BP=AD′
BC，
That is 5 − X
X=4
6，
X = 3,
Pb = 3
Therefore, D
As shown in the figure, in the pyramid p-abcd, ∠ DAB = ∠ ABC = 90 ° PA ⊥ plane ABCD, point E is the midpoint of PA, ab = BC = 1, ad = 2 Verification: (1) plane PCD ⊥ plane PAC (2) Be ‖ plane PCD
(1) Idea: find a line on the surface PCD, vertical plane PAC, and lock the segment CD after observation
On the plane ABCD, it is easy to prove CD ⊥ AC
From PA ⊥ plane ABCD, CD ⊥ PA is obtained
So CD ⊥ PAC,
So PCD ⊥ PAC
(2) Idea: find a line segment parallel to be on the surface PCD and observe the midpoint of PD after be translation
Let f be the midpoint of PD,
In triangle PAD, the bottom AD of median line EF//, and EF=AD/2=1
And AD / / BC, BC = 1, so EF / / BC and EF = BC, BCFE are parallelogram, so be / / FC is obtained
FC is on plane PCD, so be / / plane PCD
As shown in the figure, in △ ABC, point D is on edge AC, DB = BC, point E is the midpoint of CD, and point F is the midpoint of ab (1) Verification: EF = 1 2AB； (2) Pass through point a as Ag ‖ EF, intersect the extension line of be at point G, and verify: △ Abe ≌ △ age
It is proved that: (1) connect be, (1 point) ∵ DB = BC, point E is the middle point of CD, ᚉ be ⊥ CD. (2 points) ∵ point F is the middle point on the hypotenuse of RT ⊥ Abe, ᙽ EF = 12ab; (3 points) (2) [method 1] in △ ABG, AF = BF, Ag ∥ EF, EF is the median line of △ ABG, be = eg. (3 points) in △ Abe and
As shown in the figure, in △ ABC, point D is on edge AC, DB = BC, point E is the midpoint of CD, and point F is the midpoint of ab. (1) verification: EF = half ab As shown in the figure, in △ ABC, point D is on edge AC, DB = BC, point E is the midpoint of CD, and point F is the midpoint of ab. (1) verification: EF = half ab
prove:
Connect be
Because BD = BC, the triangle BDC is an isosceles triangle
Because e is the midpoint of CD, be ⊥ CD
So the triangle Abe is a right triangle
F is the midpoint of the hypotenuse ab
Because the center line of the hypotenuse of a right triangle is equal to half of the hypotenuse
EF = AB/2
As shown in the figure, in the triangle ABC, point D is on the edge AC, DB = BC, point E is the midpoint of CD, and point F is the midpoint of ab. it is proved that EF = 1 / 2Ab Please help me solve it as soon as possible,
If BC = BD, e is the midpoint of CD, then be ⊥ CD
In a right triangle Abe, EF is the center line on the hypotenuse ab
So: EF = 1 / 2Ab
As shown in the figure, in △ ABC, point D is on edge AC, DB = BC, point E is the midpoint of CD, and point F is the midpoint of ab (1) Verification: EF = 1 2AB； (2) Pass through point a as Ag ‖ EF, intersect the extension line of be at point G, and verify: △ Abe ≌ △ age
0
0
It is proved that: take point E on the extension line of Ba, make AE = AC, and connect DC and de
∵ ad bisection ∵ CAE
∴∠EAD＝∠CAD
∵AE＝AC、AD＝AD
∴△AED≌△ACD
∴DE＝DC
∵ in △ DBE:
BE＜DB+DE,
BE＝AB+AE＝AB+AC
∴AB+AC＜DB+DC
If you are satisfied with my answer, please click the "accept as satisfied" button below
If you have any other questions, you can ask our team for help
o(∩_ ∩)o
RELATED INFORMATIONS
- 1. In △ ABC, ∠ ACB = 100 °, AC = BC, point D is on AB, and BD = BC, point E is on AC, and AE = ad, EF ⊥ DC is on f 1. Find the degree of ∠ def 2: If AC = BC is removed from the title, what about the degree of ∠ def 3: If the body weight ∠ ACB = 100 ° is changed to ∠ ACB ＞∠ a, other things remain unchanged, what is the relationship between ∠ def and ∠ ACB
- 2. In the triangle ABC, the angle ACB = 90 degrees, ad bisects the angle BAC, De is perpendicular to e, EF intersects AC to F, and EC intersects ad to o
- 3. As shown in the figure, △ ABC is an isosceles triangle, ∠ ACB = 90 ° and the midpoint D of BC is de ⊥ AB, and the perpendicular foot is e to connect CE. Find the value of sina ∠ ace
- 4. In the triangle ABC, the bisector of angle B and the bisector of angle c intersect at point DDG / BC, and acab is proved to be GF = bg-cf at two points of FG
- 5. As shown in the figure: in isosceles RT △ ABC, ∠ C = 90 degrees, AC = 8, f is the midpoint of AB edge, points D and e move on AC and BC sides respectively, and keep ad = CE, connect We need the whole proof process. Thank you
- 6. [function] point P is an image of inverse scale function and positive scale function y = - 2x. PQ is perpendicular to X axis, and the coordinate of perpendicular foot is (2,0) If point m is on the inverse scale function image and the area of △ MPQ is 6, find the coordinates of point M Missing two words Point P is the intersection of an image with an inverse scale function and a positive scale function y = - 2x
- 7. The graph of the known quadratic function y = AX2 + BX + C (a ≠ 0) is shown in the figure ① ABC > 0; ② B < A + C; ③ 4A + 2B + C > 0; ④ 2C < 3B; ⑤ a + b > m (am + b) (m ≠ 1 real number) The correct conclusion is () A. 2 B. Three C. Four D. Five
- 8. As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained
- 9. D is the middle of the edge BC of the triangle ABC, De is vertical AC, DF is vertical AB, the vertical feet are e, F, and BF = CE, judge the shape of triangle ABC Is it an isosceles right triangle? Give proof
- 10. As shown in the figure, in RT △ ABC, ab = AC, ∠ a = 90 °, point D is any point of BC, DF ⊥ AB is in F, de ⊥ AC is in E, M is the midpoint of BC. Try to judge what shape △ MEF is and prove your conclusion
- 11. As shown in the figure, the vertex angle of the isosceles △ ABC is 50 ° and ab = AC. taking AB as the diameter, make a semicircle and intersect BC at point D and AC at point E BD、 De and The degree of the center angle to which AE is directed
- 12. Let d be a point on the edge BC of △ ABC, and AB2 = ad2 + BD × DC
- 13. As shown in the figure, △ ABC is an equilateral triangle, and points D, e and F are points on the segments AB, BC and Ca respectively, (1) If ad = be = CF, is △ def an equilateral triangle? Try to prove your conclusion; (2) If △ DEF is an equilateral triangle, is ad = be = CF true? Try to prove your conclusion
- 14. As shown in the figure, in the triangle ABC, ∠ 1 = ∠ 2, G is the midpoint of AD, lengthening BG intersecting AC with E.F is a point on AB, CF ⊥ ad is in H (1) Ad is the angular bisector of the triangle Abe; (2) Be is the center line on the side ad of the triangle abd; (3) Ch is the height on the edge ad of the triangle ACD A. 1 B. 2 C. Three D. 0
- 15. It is known that, as shown in the figure, △ ABC is an equilateral triangle, AE = CD, BQ ⊥ ad at Q, be crossing ad at point P, Confirmation: BP = 2pq
- 16. It is known that, as shown in the figure, in △ ABC, ad and AE are the bisectors of the height and angle of △ ABC respectively, if ∠ B = 30 ° and ∠ C = 50 ° (1) Find the degree of ∠ DAE; (2) What is the relationship between ∠ DAE and ∠ C - ∠ B? (no need to prove)
- 17. It is known that AD and AE are the angular bisector and height of triangle ABC respectively, ∠ B = 50 °, ead = 10 ° and ∠ C=____
- 18. As shown in the figure, in △ ABC, the angular bisectors ad, be and CF intersect at point h, passing through point h as Hg ⊥ AC, and the foot perpendicular to g, then ∠ ahe = ∠ CHG? Why?
- 19. As shown in the figure, △ ABC is an equilateral triangle, and ad = be = CF, is △ def an equilateral triangle? Write it out in 20 minutes,
- 20. As shown in the figure, in △ ABC, AC = BC, ∠ ACB = 90 °, D is a point in △ ABC, ∠ bad = 15 °, ad = AC, CE ⊥ ad in E, and CE = 5 (1) Find the length of BC; (2) Confirmation: BD = CD