![](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, △ ABC is an equilateral triangle, and ad = be = CF, is △ def an equilateral triangle? Write it out in 20 minutes,
Although there is no figure, because: △ ABC is an equilateral triangle, ab = BC = AC, angle a = angle B = angle c = 60 degrees, and then prove that the triangle ADF is all equal to the triangle, bed is all equal to the triangle CFE, so DF = de = EF, so △ DEF is an equilateral triangle
As shown in the graph, D, e and F are points on each edge respectively, and ad = be = CF Verification: △ DEF is an equilateral triangle
It is proved that ∵ △ ABC is an equilateral triangle and ad = be = CF
∴AF=BD=CE，
And ? a = ∠ B = ∠ C = 60 °,
∴△ADF≌△BED≌△CFE（SAS），
∴DF=ED=EF，
The △ DEF is an equilateral triangle
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
(1) Delta DEF is an equilateral triangle
The proof is as follows:
∵ △ ABC is an equilateral triangle,
∴∠A=∠B=∠C，AB=BC=CA，
And ∵ ad = be = CF,
ν DB = EC = FA, (2 points)
≌ △ ADF ≌ △ bed ≌ △ CFE, (3 points)
/ / DF = de = EF, that is, △ DEF is an equilateral triangle; (4 points)
(2) Ad = be = CF holds
The proof is as follows:
As shown in the figure, ∵ △ DEF is an equilateral triangle,
∴DE=EF=FD，∠FDE=∠DEF=∠EFD=60°，
∴∠1+∠2=120°，
ABC is an equilateral triangle,
∴∠A=∠B=∠C=60°，
∴∠2+∠3=120°，
Ψ 1 = ∠ 3, (6 points)
Similarly, ∠ 3 = ∠ 4,
≌ △ ADF ≌ △ bed ≌ △ CFE, (7 points)
/ / ad = be = cf. (8 points)
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
0
0
Make de ⊥ AB in E
∵ ad is ᙽ ABC angle bisector, and ᙽ C = 90 ° and ∵ AED = 90 °
∴CD=ED
Ad = ad
∴△ADC≌ADE（HL）
AC = AE
In △ ABC and △ BDE
∠C=∠BED,∠B=∠B
∴∠BAC=∠BDE
∵ △ ABC is an isosceles triangle
∴∠BAC=∠B
∴∠BDE=∠B
De = be
∴AB=AE+BE=AC+DE=AC=CD
It is known that ad is the bisector of the base angle of the isosceles triangle ABC. The angle c = 90 degrees. AB = AC + CD? It's an isosceles right triangle
The answer is ab = AC + CD. It is proved that if the angle CAE, ad is its bisector, C is a right angle, AED is a right angle. Therefore, CD = de (can also be obtained by the theorem of "the distance between the point on the bisector and the two sides of the angle is equal") AC = AE triangle ABC is an isosceles right angle
It is known that ad is the bisector of ABC base angle of isosceles triangle, ∠ C = 90 °, can you prove AB = AC + CD? Thank you
Extend AC to e so that CE = CD
Δ DCE is an isosceles right triangle, so ∠ CED = 45 °
Therefore, CED = abd = 45 degrees
Because ∠ ead = ∠ bad, ad is a common edge,
So △ ead is all equal to △ bad
AB=AE
AB = AC + CE
Because CE = CD
So AB = AC + CD
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?
Reason: ∵ ad, be and CF are the angular bisectors of ad, be, CF are the angular bisectors of △ ABC, bad = ∠ CAD = x, ∠ Abe = ∠ CBE = y, ∠ BCF = ∠ ACF = Z, then 2x + 2Y + 2Z = 180 ° i.e. x + y + Z = 90 ° in △ AHB, ? ahe is the outer angle of △ AHB, ? ahe = ∠ bad + ∠ Abe = x + y = 90 ° - Z
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?
∠AHE=∠CHG．
Reason: ∵ ad, be, CF are the angular bisectors of ᙽ ABC,
ν = Abe = BAE,
Then 2x + 2Y + 2Z = 180 °,
That is, x + y + Z = 90 °,
In △ AHB,
∵ ahe is the outer angle of AHB,
∴∠AHE=∠BAD+∠ABE=x+y=90°-z，
In △ CHG, ∠ CHG = 90 ° - Z,
∴∠AHE=∠CHG．
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?
∠AHE=∠CHG．
Reason: ∵ ad, be, CF are the angular bisectors of ᙽ ABC,
ν = Abe = BAE,
Then 2x + 2Y + 2Z = 180 °,
That is, x + y + Z = 90 °,
In △ AHB,
∵ ahe is the outer angle of AHB,
∴∠AHE=∠BAD+∠ABE=x+y=90°-z，
In △ CHG, ∠ CHG = 90 ° - Z,
∴∠AHE=∠CHG．
RELATED INFORMATIONS
- 1. 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?
- 2. It is known that AD and AE are the angular bisector and height of triangle ABC respectively, ∠ B = 50 °, ead = 10 ° and ∠ C=____
- 3. 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)
- 4. 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
- 5. 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
- 6. 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
- 7. Let d be a point on the edge BC of △ ABC, and AB2 = ad2 + BD × DC
- 8. 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
- 9. 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
- 10. 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
- 11. 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
- 12. It is known that in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at point D, be bisection ∠ ABC, intersection ad at point m, an bisection ∠ DAC, intersection BC at point n Verification: the quadrilateral amne is rhombic
- 13. As shown in the figure, in △ ABC, ad bisects ∠ BAC to BC to D, be ⊥ AC to e, and ad to F. it is proved that ∠ AFE = 1 2（∠ABC+∠C）．
- 14. Let us know that the triangle ABC, D is a point on BC, e is a point on AC, and BD = CE, connect ad, BD, intersect with point F, and find the degree of angle AFE
- 15. 0
- 16. As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC is at point D, de ∥ AB intersects AC at point E. is △ ade an isosceles triangle? Why?
- 17. As shown in the graph, △ ABC, ab = AC, points D, e, f are on edges BC, AB, AC respectively, and BD = CF, ∠ EDF = ∠ B, is there any triangle congruent with △ BDE in the graph? And explain the reason
- 18. As shown in the figure, points a, B and E are on the same straight line, △ ABC and △ BDE are equilateral triangles. Ad intersects BC at f and CE intersects BD with AD at g and h respectively
- 19. As shown in the figure, △ ABC is an isosceles right triangle, ∠ ACB = 90 ° and ad is the center line on the side of BC. Through C, the vertical line of ad is made, which intersects AB at point E and ad at point F. it is proved that ∠ ADC = ∠ BDE
- 20. In the triangle ABC, points D, e, f are the midpoint of sides BC, Ca and ab respectively. Then AB + AD + BC + be + CF (all vectors)=