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Ad is the height of the triangle ABC, and the circle O intersects with AD as its diameter AB.AC At point E. F, it is proved that angle B is equal to angle AFE
As auxiliary line: connect ed
Then the triangle AED is a right triangle
Since ad is perpendicular to BC, the triangle ADB is also a right triangle
So angle ade = angle B
The angle ade and angle AFE correspond to the same arc AE
So angle ade = angle AFE
Angle B = angle AFE
In the triangle ABC, ad is the center line on the side of BC, e is a point on AC, be and ad intersect F. if the angle FAE = angle AFE, it is proved that AC = BF In the triangle ABC, ad is the center line on BC side, e is a point on AC, be and ad intersect F If angle FAE = angle AFE Verification: AC = BF
It is proved that CG ∥ be is made, ad is delivered to g, and BG is connected
∵CG‖BE
∴∠BFD=∠CGD,∠FBD=∠GCD
And BD = CD
∴△BFD≌△CGD
BF=CG
∵CG‖BE
∴∠AFE=∠AGC
And ∠ AFE = ∠ FAE
∴∠AGC=∠FAE
∴AC=GC
Proved BF = CG
∴AC = BF
As shown in the figure, in the triangle ABC, ∠ ACB = 90 degrees, AC = BC, be is perpendicular to CE at point E, ad is perpendicular to CE at D, and ad-be = De
prove:
∵BE⊥CE,AD⊥CE
∴∠BEC＝∠ADC＝90
∴∠BCE+∠CBE＝90
∵∠ACB＝90
∴∠BCE+∠ACD＝90
∴∠CBE＝∠ACD
∵AC＝BC
∴△ACD≌△CBE （AAS）
∴BE＝CD,AD＝CE
∵DE＝CE-CD
∴DE＝AD-BE
Hope to get your adoption,
As shown in the figure, in △ ABC, ∠ ACB = 90 °, AC = BC, be ⊥ CE at point E. ad ⊥ CE at point D Verification: △ BEC ≌ △ CDA
It is proved that ∵ be ⊥ CE in E, ad ⊥ CE in D,
∴∠BEC=∠CDA=90°，
In RT △ BEC, ∠ BCE + ∠ CBE = 90 °,
In RT △ BCA, ∠ BCE + ∠ ACD = 90 °,
∴∠CBE=∠ACD，
In △ BEC and △ CDA, ∠ BEC = CDA, ∠ CBE = ∠ ACD, BC = AC,
∴△BEC≌△CDA．
As shown in the figure, in the triangle ABC, the angle ACB = 90 degrees, D is the midpoint of BC, De is vertical BC, CE / / AD, if AC = 2, CE = 4, find the circumference of quadrilateral aceb
We can see that AC / / De, AD / / CE, so aced is a parallelogram, so de = AC = 2,
From Pythagorean theorem, we can get CD ^ 2 = CE ^ 2-DE ^ 2; therefore, there is CD = 2 times the root sign 3
So CB = 4 times root 3
AB^2=AC^2+CB^2
As shown in the figure, it is known that ad is the bisector of the inner angle of the triangle ABC. It is proved that AB / AC = BD / CD
This is the angular bisector theorem
Using the positive metaphysical theorem
AB/sin∠ADB=BD/sin∠BAD (1)
AC/sin∠CDB=CD/sin∠CAD (2)
Ad is an angular bisector, sin ∠ bad = sin ∠ CAD
∠ADB+∠CDB=180
sin∠ADB=sin∠CDB
(1) Formula (2)
AB/AC=BD/CD
As shown in the figure, in the known triangle ABC, ad is the bisector of the angle, and the point E is on AB [AB > AC], and AE = AC, connected with CE. If the angle ACB = 68 degrees, the angle B = 54 degrees. Try to find the angle
Let BCE be x, then AEC is 54 + X. in triangle AEC, BAC is known to be 58, and equation 180 = (54 + x) * 2 + 58, the solution is 7
It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°
In this paper, we have the idea that ᙽ EBD = x
As shown in the figure, in the triangle ABC, the angle c = 2, the angle B, ad is the angular bisector of the triangle ABC, the angle EDB = the angle B, and ab = AC + CD
Angle B is equal to angle BDE, so be = De, angle DEA = angle B + angle BDE = 2 angle B = angle C
Ad is the bisector of an angle, so the angle CAD = angle DAB, so the three angles of the triangle ACD and ade are equal, and one edge coincides, so it is congruent. CD = De
AB=AE+BE=AC + DE =AC+DE
It is known that the figure ad, be and CF are the bisectors of the equilateral triangle ABC. Find that the triangle DEF is an equilateral triangle
Suppose that ab = BC = CA = 2x, and ad, be, CF are the three angular bisbisectors of the equilateral △ ABC, then ad, be, CF and ad, be, and CF are vertical bisectors of the three sides of the equilateral △ ABC, i.e., D, e, f are the midpoint of the three sides, so EF / / BC, Ed / / AB, DF / / AC, EF / / BC, Ed / / AB, DF / / AC, because ∠ a = ∠b = C = 60 ° and EF / / BC, so ∠afe = ∠ B = 60 ° so ∠ AEF = 180 ° -, a - ∠ AFE = 180-60-60 = 60 °, that △ AEF is equal, that is, △ AEF is equal equal equal equal equal equal equal to 180 ° - 60-60-60 side triangle, Moreover, e and F are the midpoint of AC and ab respectively, so AE = AF = EF = x; by analogy, it is proved that FD and ED are both X, that is, △ DEF is an equilateral triangle
RELATED INFORMATIONS
- 1. As shown in the figure, it is known that ad is the bisector angle of the isosceles triangle ABC, and the angle c = 90 degrees proves that ab = AC + CD
- 2. 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 °, calculate the degree of ∠ DAE
- 3. As shown in the figure, in △ ABC, it is known that ab = BC = Ca, AE = CD, ad and be intersect at point P, BQ ⊥ ad at point Q. verification: BP = 2pq
- 4. It is known that in the triangle ABC, D is a point on AB, be is perpendicular to ad, CF is perpendicular to ad, and perpendicular feet are e and f respectively If ad is the central line of the triangle ABC, be = CF is proved If ABC is a triangle of ABC, prove that it is a triangle fast
- 5. Let d be a point on the edge BC of △ ABC, and AB2 = ad2 + BD × DC
- 6. As shown in the figure, in the isosceles △ ABC, ab = AC. the semicircle with diameter AB intersects BC at point D and AC at point E De = 40 °, find ∠ A and Degree of AE
- 7. As shown in the figure, in the right angle trapezoid ABCD, ad ‖ BC, ∠ ABC = 90 °, BD ⊥ DC, BD = DC, CE bisection ∠ BCD, crossing AB at point E, crossing BD at point h, en ∥ DC intersecting BD at point n ①BH=DH；②CH=( 2+1)EH；③S△ENH S△EBH＝EH EC. The correct one is () A. ①②③ B. Only ② and ③ C. Only two D. Only 3
- 8. As shown in the figure, in the triangle ABC, BC = 6, e and F are the midpoint of AB and AC respectively, the moving point P is on the ray EF, BP intersects CE at D, bisector of angle CBP intersects CE at Q, when CQ = 1 / 3 * ce, EP + BP=____ .
- 9. As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are on BC and AC respectively, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles triangle As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = AC, points D and E are respectively on BC and AC, and BD = CE, M is the midpoint of AB, and △ MDE is isosceles right triangle. Please explain the reasons
- 10. As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD
- 11. As shown in the figure, in △ ABC, the angle BAC = 90 degrees, ad ⊥ BC bisects ∠ ABC at point D and BF, then △ AEF is an isosceles triangle? And the proof is given
- 12. In the equilateral triangle ABC, ad and CE are the heights on BC and ab respectively, and AD and CE intersect at point F, what is the degree of ∠ AFE? If points D and e run on BC and ab respectively, in order to make the conclusion (1) hold, please guess what quantitative relationship BD and AE should satisfy and give proof
- 13. It is proved that ab minus AC is greater than BD minus DC
- 14. As shown in the figure, in the triangle ABC, ad is perpendicular to BC, CE is perpendicular to AB, BC = 6cm, ab = 8cm, calculate the value of AD: CE
- 15. As shown in the figure, in triangle ABC, angle B is equal to angle c, points D, e, f are on AB, BC and AC respectively, and BD = CE, angle def = angle B, whether there is and triangle BD in the graph
- 16. As shown in the figure, B is a point on the line ad, △ ABC and △ BDE are equilateral triangles, which connect CE and extend the extension line crossing ad at point F, and the circumscribed circle ⊙ o of ⊙ ABC intersects CF at point P (1) It is proved that be is the tangent of ⊙ o; (2) If CP = 2, PF = 8, find the length of AC
- 17. Let D, e, f be the midpoint of BC, AC and ab of the triangle ABC respectively How to do such a topic,
- 18. In the triangle ABC, we know that D is a point on the edge of ab. if vector ad = 2 vector dB, vector CD = 1 / 3 vector Ca + X vector CB, then x is equal to A.2/3 B.1/3 C.-1/3 D.-2/3
- 19. In the triangle ABC, if the angle BAC = 120 degrees, ab = 2, AC = 1, D is a point on the edge BC, DC = 2bd, then the vector ad points multiply the vector BC =?
- 20. Given that the circumference of the right triangle ABC is 4 + 4, the root sign 2, and the length of the center line on the hypotenuse is 2, then the area of the triangle is?