![](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>)
As shown in the figure, points D and E are on AB and AC respectively, and ∠ ABC = ∠ AED. If de = 4, AE = 5, BC = 8, then the length of AB is______ .
In △ ABC and △ AED,
∵∠ABC=∠AED，∠BAC=∠EAD，
∴△AED∽△ABC，
∴AB
AE=BC
ED，
And ∵ de = 4, AE = 5, BC = 8,
∴AB=10．
So the answer is: 10
As shown in the figure, ab = AE, ∠ 1 = ∠ 2, ∠ C = ∠ D Verification: △ ABC ≌ △ AED
It is proved that: ∵ 1 = ∠ 2,
∴∠1+∠EAC=∠2+∠EAC，
That is ∠ BAC = ∠ EAD,
∵ in △ ABC and △ AED,
∠D＝∠C
∠BAC＝∠EAD
AB＝AE ，
∴△ABC≌△AED（AAS）．
As shown in the figure, D and E are the points on the edge AB and AC of △ ABC, BD / ad = AE / CE = 3, and ∠ AED ∠ B, then the area ratio of △ AED to △ ABC is
Δ ABC increases BF based on AC
Δ AED with AE as the base, sitting height DG
Because ∠ AED = ∠ B ∠ a = ∠ a
So △ AED and △ ABC are equilateral triangles
Because BD / ad = AE / CE = 3
So BF / DG = 4
AC/AE=4/3
Δ AED area = 1 / 2DG * AE
Δ ABC area = 1 / 2BF * AC
(1/2DG*AE)/(1/2BF*AC)=3/16
Benefactor, I see your bones are very clear,
They are elegant and have their own wisdom,
He is a wonder in the Wulin
If you concentrate on your studies, you will become great in the future,
I have a little test, please click next to the answer
"Selected as satisfactory answer"
In the right triangle ABC, ∠ C + 90 ° and D.E are points on AC and ab respectively, and ad = BD, AE = BC, then de ⊥ ab. please explain the reason~ Please answer carefully It's the reason!
Because triangle ABC is a right triangle, triangle BCD is a right triangle
Because ad = BD, the triangle abd is an isosceles triangle. The angle BAC = 30 degrees, the angle DBC = 30 degrees (easy to prove)
Because ad = BD, AE = BC
So triangle BCD = triangle ade, so De is vertical ab
The idea is right. I can hardly type it out. You can sort it out yourself
It is known that in △ ABC, ab = AC, ad, AE bisect ∠ BAC and ∠ CAF respectively, and ad intersects BC at point DAE = DC. It is proved that the quadrilateral adce is a rectangle
It is proved that: ∵ AB = AC, ad bisection ∵ BAC ᙽ ① ad ⊥ BC (three lines in an isosceles triangle) 2 DAC = half ? BAC ? AE bisection caf ? EAC = half ? caf ? BAC + ? CAF = 180 ° ? DAC + ∵ EAC = 90 °, AE / / AE / / DC ? DC ? AE = DC
In △ ABC, ∠ BAC = 90 ° and ad is the height of BC. The bisector of ∠ ABC intersects ad at e, EF ‖ BC and AC at point F. can you guess the quantitative relationship between segment AE and CF? Please state the reasons
AE = CF, reason: lengthen Fe to meet AB to g. make GH ∥ AC to BC to H. connect eh, ∵ EF ∥ BC, ? FG ∥ BC. ᙽ quadrilateral fghc is a parallelogram, GH = CF. FCD = ∠ EGH, ? BAC = 90 °, ad is the height on the edge of BC, ACB = ∵ bad, ? EGH = ∥ bad, ∵ FG ∥ BC, ad ⊥ BC
Ad ⊥ BC, BD = DC, C on the vertical bisector of AE, what is the relationship between the length of AB, AC and CE? As shown in the figure, ad ⊥ BC, BD = DC, C on the vertical bisector of AE, what is the relationship between the length of AB, AC and CE? To write out the whole process, the picture comes to me space, who will not enter the space?
AB = AC = CE. Prove: ad ⊥ BC, BD = DC, deduce AB = AC (if you can't think clearly, we can use the congruent proof. In the right triangle abd and ADC, ad = ad, BD = DC, we can know that the two triangles are congruent, so AB = AC)
C is on the vertical bisector of AE, and the distance from the point on the vertical bisector of the line segment to the two ends of the line segment is equal
Ad vertical BC, BD is equal to DC, point C is on the vertical bisector of AE, what is the relationship between the length of AB, AC, CE? What is the relationship between AB, BD and de?
From the title "ad vertical BC, BD equals DC", it can be seen that ad is the vertical bisector of BC (median vertical line)
AB = AC is obtained from the mid perpendicular theorem
Similarly, BC is the vertical line of AE
So AC = CE
In conclusion, ab = AC = CE
Is the relationship between AB, BD and de confirmed by the owner?
Ad vertical BC, BD = DC, point C on the vertical bisector of AE, what is the relationship between the length of AB, AC, CE? What is the relationship between ab + BD and de
AB=AC=CE
AB+BD>DE
Answer: suppose that the intersection point of BC and ad is f, BD = DC, DF is vertical BC, BF = FC is obtained, so AB = AC is deduced
Similarly, C is on the vertical bisector of AE, so ace is an isosceles triangle, i.e. AC = CE, thus AB = AC = CE
AB + BD = AC + DC = CE + DC according to the triangle theorem, the sum of two sides is greater than the third side, so DC + CE > de is ab + BD "de"
As shown in the figure, in ladder ABCD, EF is parallel to ad, parallel to BC, AE = 3, EB = 7, DC = 13, then FC=
FC=7/10*13=9.1
RELATED INFORMATIONS
- 1. In trapezoid ABCD, AD / / BC, ad = 3, BC = 5, points E and F are on AB and DC respectively, EF / / BC, if AE: EB = 2:3, find EF I also calculated 3. But the answer is 2
- 2. In △ ABC, D is a point on the edge of BC, BD = 1 2dc, ∠ ADB = 120 ° ad = 2, if the area of △ ADC is 3, then AB = () A. 1 B. Five C. Seven D. 2 Two
- 3. Given the triangle ABC, if ∠ C = 90 °, the square of 4B + the square of 3C = four times the root sign 3bC, find the value of cosa
- 4. In the right triangle ABC, ∠ C = 90 °, BC: AC = 1: radical 3, CD ⊥ AB in D, find s triangle CDB: s triangle ABC
- 5. In the RT triangle ABC, the angle ACB = 90 degrees, ab = 4, take AC and BC as the diameters to make semicircles, and the areas are denoted as S1 and S2 respectively?
- 6. In the RT triangle ABC, ∠ C = 90 °, ab = radical 10, AC: BC = 2:1, calculate the perimeter and area of RT triangle
- 7. As shown in △ ABC, the angle ACB = 90 °, the angle a = 60 ° and the height of the hypotenuse CD = the root sign 3. Find the length of ab D is on AB, CD is perpendicular to AB, AC is perpendicular to CB
- 8. In RT △ ABC, ∠ C = 90 ° s = 18 root sign 3, a
- 9. In the triangle ABC, if edge B + C = radical 2 + 1, angle c = 45 degrees, angle B = 30 degrees, what are the values of edges B and C
- 10. As shown in the figure, △ ABC, ab = AC, CD bisection ∠ ACB, intersection AB at point D, ∠ BDC = 150 ° and find the degree of ∠ a
- 11. In △ ABC, a = 2, B = 2. If the triangle has a solution, what is the value range of a? A belongs to (0 degree, 90 degree). Why not?
- 12. As shown in the figure, triangle ABC is all equal to triangle ade, if angle BAE = 120 ° angle bad = 40 ° angle BAC =?
- 13. As shown in the figure AB = ad ∠ bad = ∠ CAE AC = AE, ab = ad
- 14. As shown in the figure, it is known that AE is the center line of triangle ABC, and the perimeter of triangle Abe is 2cm more than that of triangle ace, if AB + A If AB + AC = 18cm, request AB, AC length
- 15. The known points D and E are on the sides AB and AC of the triangle ABC, ab = 10cm, AC = 6cm, ad = 2cm=__ Triangle ade is similar to triangle ABC Since there is no drawing, there will be two situations I have one more question to answer
- 16. In the triangle ABC, D is a point on the edge of AC, the angle DBC = angle a, BC = radical 6, AC = 3, find the length of CD Urgent``````
- 17. As shown in the figure, in the RT triangle ABC, the angle c = 90 degrees, the angle ABC = 30 degrees, the bisecting angle BAC BD, and the parallel AC prove that AE = be As shown in the figure, the angle c = 90 degree angle ABC = 30 degree ad bisection angle BAC BD parallel AC (1) proves AE = be (2) Verify BC + CE = de
- 18. As shown in the figure, in triangle ABC, angle ACB = 90 degrees, AE = AC, BC = BD, calculate the degree of angle ECD Ed on ab
- 19. As shown in the figure, in △ ABC, ad is an angular bisector, e is a point on ad, and CE = CD. If AE = 2DE, the area of △ ace is 6, calculate the area of △ ABC
- 20. In RT △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of BC, e and F are on AC and AB, and AE = BF. It is proved that △ DEF is an isosceles right triangle