![](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 AD and AE are the angular bisector and height of triangle ABC respectively, ∠ B = 50 °, ead = 10 ° and ∠ C=____
Angle c = 70 or 30 degrees
It is known that in △ ABC, ∠ C > b, AE are the angular bisectors of the triangle ABC, and ad ⊥ BC is in D. try to explain ∠ ead = 1 / 2 (∠ C - ∠ b)
prove:
∵ BAC = 180 - (∠ B + ∠ C), AE bisect ∠ BAC
∴∠CAE＝∠BAC/2＝90-（∠B+∠C）/2
∵AD⊥BC
∴∠CAD+∠C＝90
∴∠CAD＝90-∠C
∴∠EAD＝∠CAE-∠CAD
＝90-（∠B+∠C）/2-90+∠C
＝（∠C-∠B）/2
In this paper, we prove that the angle of a triangle is equal to the angle of a B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B, B
It is proved that: take point E on AC, make AE = AB, connect de ? ad bisection ∵ ad bisection ∵ BAC ∵ bad ∵ CAD ? ab ? CAD ∵ ab ⊙ CAD ⊙ CAD ∵ ab ? CAD ? ad ? ad CE ν BD = CE ∵ AC = AE + ce
If the angle of B is equal to the angle of B, the proof is proved In a hurry ~ ~ ~ ~ ~ ~ ~ hand in tomorrow! speed
It is proved that taking a point E on AC makes AE = ab
∵AC=AB+BD
∴CE=AC-AE=AC-AB=BD
In triangle abd and triangle ade
∵ ad is the angular bisector of the triangle ABC
∴∠BAD=∠DAE,
AB = AE, ad is the common edge
≌ triangle ADB ≌ triangle ade (edge, angle, edge)
So BD = De,
∠B=∠AED ①
CE = BD
Thus CE = De
Then ∠ EDC = ∠ C
It is concluded that ∠ AED = ∠ EDC + ∠ C = 2 ∠ C ②
From ① and ②, we get ∠ B = 2 ∠ C
In △ ABC, ad is the height of BC edge, ∠ B = 2 ∠ C. proof: CD = AB + BD
It is proved that DM = DB is cut on DC,
∵AD⊥BC，DM=BD，
The ad is the vertical bisector of BM,
ν AB = am (the distance from the point on the vertical bisector of the line segment to both ends of the line segment is equal),
Ψ B = ∠ AMB (equilateral and equal angle),
∵∠B=2∠C，∠AMB=∠C+∠MAC，
∴∠MAC=∠C，
∴AM=CM，
∴CM=AB，
∴CD=DM+MC=BD+AB．
In the triangle ABC, ad bisects the angle BAC, and the angle B is equal to the angle C. It is proved that ab + BD = AC right off. Immediately.
It is proved that finding a point E on AC makes AE = ab
Ad bisects angle BAC, angle bad = angle DAE, ad = ad, AE = AB, so triangle abd is equal to triangle AED
BD＝DE
Angle B = angle AED = 2 angle c = angle c + angle EDC
Angle EDC = angle c
DE＝EC＝BD
So AB + BD = AE + EC = AC
As shown in the figure, ad is the height of triangle ABC, angle B is equal to angle C. It is proved that CD = AB + BD by axisymmetric property
As shown in the figure, point E is the symmetric point of point C with respect to ad, then AE = AC, ∠ e = ∠ C, because ∠ ABC = 2 ∠ C = 2 ∠ e, so ∠ EAB = ∠ e, so AB = EB, so AB + BD = ed, we know from the symmetry that ED = &
In △ ABC, ad is the height of BC edge, ∠ B = 2 ∠ C. proof: CD = AB + BD
0
0
Hello, please click "accept as the answer", Connect ed because: CD is the bisector of ∠ ACB, so: Triangle ace is isosceles triangle AC = CE, because: CD is vertical bisector AE, so: Triangle AED is isosceles triangle ad = De, because: Triangle ACD is equal to triangle ECD, so: ∠ CAD = ∠ CED and ∠ a = 2 ∠ B, so ∠ B = ∠ BDE, so: de = be BC = be + CE = AC + de = AC + ad That is: BC = AC + ad
As shown in the figure, in the triangle ABC, angle a = 2, angle B, CD is the bisector of angle ACB. It is proved that BC = AC + ad
Proof: extend CA to e, make AE = ad, connect ed
∵AE=AD,∴∠E=∠ADE,
∴∠CAD=∠E+∠ADE=2∠E,
∵∠CAD=∠2∠B
∴∠E=∠B,∠ECD=∠BCD,AD=AD
∴△ECD≌△BCD
∴BC=EC=AC+AE=AC+AD
RELATED INFORMATIONS
- 1. 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)
- 2. 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
- 3. 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
- 4. 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
- 5. Let d be a point on the edge BC of △ ABC, and AB2 = ad2 + BD × DC
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
- 11. 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?
- 12. 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,
- 13. 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
- 14. 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
- 15. 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）．
- 16. 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
- 17. 0
- 18. 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?
- 19. 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
- 20. 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