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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）．
∵ the sum of the inner angles of the triangle is 180 degrees,
∴∠BAC=180°-（∠ABC+∠C），
∵ ad bisection ∵ BAC to BC to d,
∴∠DCA=1
2∠BAC=90°-1
2（∠ABC+∠C），
∵ be ⊥ AC to E,
∴∠AFE=90°-∠FAE=90°-90°+1
2（∠ABC+∠C）=1
2（∠ABC+∠C）．
If angle DAC = angle B, CD = CE, then ace of triangle is similar to bad
prove:
Because ∠ B = ∠ DAC
CE = CD
So ∠ ADC = ∠ CED
Also ∠ CED = ∠ DAC + ∠ ECA
∠CDA=∠B+∠BAD
So ∠ ECA = ∠ bad
So △ ace is similar to △ bad
I will use mathematical expression to change it
I hope I can help you
Ad is the midline of △ ABC, e is the midpoint of AD, if ∠ DAC = ∠ B, CD = CE
∠edc=∠B+∠bad
∠dec=∠dac+∠aec
Because CD = CE, ∠ EDC = ∠ Dec
Because ∠ DAC = ∠ B,
Therefore, from the above, ∠ bad = ∠ AEC
Because ∠ DAC = ∠ B, two angles of △ ace and △ bad are equal, so they are similar
As shown in the figure, it is known that ad is the center line of ABC, and ∠ DAC = ∠ B, CD = CE (1) Verification: △ ace ∽ △ bad: (2) If AB = 12, BC = 8, try to find the length of AC and AD
(1) Proof: CD = CE,
∴∠CDE=∠CED，
∴∠AEC=∠BDA，
And ∵ DAC = ∠ B,
∴△ACE∽△BAD；
（2）∵∠DAC=∠B，∠ACD=∠BCA，
∴△ACD∽△BCA，
∴AC
BC=CD
AC, i.e. AC
8=4
AC，
∴AC=4
2，
∵△ACE∽△BAD，
∴AC
BA=CE
Ad is 4
Two
12=4
AD，
∴AD=6
2．
If the angle DAC = angle B, BD = CE. Try to prove that ace of triangle is similar to bad of triangle
If you draw a graph, you will find that if you want to prove that the ace of the triangle is similar to the bad of the triangle, it is enough to prove that the angle AEC = the angle BDA. If the angle AEC = the angle BDA, so the angle CED = the angle ADC (complementary), so CE = CD = dB, so the angle CAD = angle DAB
It is known that ad is the height of the triangle ABC, EA is the bisector of the angle of the triangle ABC. If the angle B = 44 degrees, the angle c = 78 degrees, the degree of the angle DAE
17 degrees
In the triangle ABC, ad and AE are the bisectors of the height and angle of the triangle ABC respectively, ∠ C = 60 ° and B = 28 ° to find ∠ DAE
ADC=90
DAC=180-60-90=30
BAC=180-28-60=92
EAC=1/2BAC=46
DAE=CAE-CAD=46-30=16
It is known that: as shown in the figure, ad is the angular bisector of the triangle ABC, e is a point on the BC extension line, ∠ EAC = ∠ B. verification: ∠ ade = ∠ DAE
It is proved that ad bisects ∠ BAC, so ∠ bad = ∠ CAD
The ∠ ade is △ bad outside angle, so ∠ ade = ∠ bad + ∠ B
∠DAE=∠CAD+∠EAC
Because ∠ B = ∠ eac
So ∠ ade = ∠ DAE
As shown in the figure, in △ ABC, ab = AC, AE is the bisector of ∠ BAC external angle ∠ DAC. Try to judge the position relationship between AE and BC, and explain your conclusion
The relationship between AE and BC is AE ∥ BC
∵AB=AC，
∴∠B=∠C，
And ? DAC = B + ∠ C = 2 ∠ C, AE is the bisector of DAC,
∴∠DAC=2∠EAC，
∴∠C=∠EAC，
/ / AE ∥ BC (the internal staggered angle is equal, and the two lines are parallel)
As shown in the figure, in △ ABC, ∠ BAC = 80 °, B = 60 °, ad ⊥ BC, the foot of perpendicularity is D, AE bisects ∠ DAC, and calculates ∠ AEC degree
∵∠B=60°，AD⊥BC，
∴∠BAD=30°，
∵∠BAC=80°，∴∠DAC=50°，
∵ AE bisection ∠ DAC, DAE = 25 °,
∴∠BAE=55°，
∴∠AEC=∠BAE+∠B=55°+60°=115°．
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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,
- 4. 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?
- 5. It is known that AD and AE are the angular bisector and height of triangle ABC respectively, ∠ B = 50 °, ead = 10 ° and ∠ C=____
- 6. 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)
- 7. 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
- 8. 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
- 9. 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
- 10. Let d be a point on the edge BC of △ ABC, and AB2 = ad2 + BD × DC
- 11. 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
- 12. 0
- 13. 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?
- 14. 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
- 15. 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
- 16. 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
- 17. 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)=
- 18. In △ ABC, it is known that BD and CF are high, M is the midpoint of BC, and N is the midpoint of DF Yes,
- 19. In the triangle ABC, D is a point on the edge of AB, and vector ad = 2 vector dB, vector CD = 1 / 3 vector Ca + n vector CB, then n What is the value of N?
- 20. Any line passing through the center of gravity of △ ABC intersects AB and AC at points D and e respectively AD＝x AB， AE＝y AC, XY ≠ 0, then 1 X+1 The value of Y is______ .