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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
Proof: ∵ ad ⊥ BC,
∴∠BDA=90°，
∵∠BAC=90°，
∴∠ABC+∠C=90°，∠ABC+∠BAD=90°，
∴∠BAD=∠C，
∵ an bisection ∠ DAC,
∴∠CAN=∠DAN，
∵∠BAN=∠BAD+∠DAN，∠BNA=∠C+∠CAN，
∴∠BAN=∠BNA，
∵ be bisection ∵ ABC,
∴BE⊥AN，OA=ON，
Similarly, OM = OE,
The quadrilateral amne is a parallelogram,
The parallelogram amne is a diamond
As shown in the figure, △ ABC, ab = AC, ∠ BAC = 120 ° and the vertical bisector EF of AC intersects AC at point E and BC at point F. verification: BF = 2cf
It is proved that: connect AF, (1 minute) ∵ AB = AC, BAC = 120 °, B = ∵ C = 180 ° − 120 ° 2 = 30 °, (1 minute)
As shown in the figure, △ ABC, ab = AC, ∠ BAC = 120 ° and the vertical bisector EF of AC intersects AC at point E and BC at point F. verification: BF = 2cf
It is proved that: connect AF, (1 minute) ∵ AB = AC, BAC = 120 °, B = ∵ C = 180 ° − 120 ° 2 = 30 °, (1 minute)
The triangle ABC, ad is the center line of BC and E is the point above AC. be and ad intersect F. if ∠ FAE = ∠ AFE, AC = BF is proved
It is proved that: extend FD to m, make DM = DF, connect cm
Then ⊿ CDM ≌ ⊿ BDF (SAS), CM = BF; m = BFD
And ∠ BFD = ∠ AFE = ∠ FAE, so ∠ FAE = ∠ m, AC = cm
So, AC = BF
As shown in the figure, in the triangle ABC, ad is the center line of BC, f is a point on AC, BF intersects with AD at point E, if ∠ FAE = ∠ AEF If ∠ FAE = ∠ AEF, find AC = be
Extend ed to point h so that DH = De
Because ad is the midline
So BD = CD
Angle BDE = angle CDH
DH=DE
It is proved that the BDE of triangle is equal to CDH of triangle
So angle bed = angle H
Because the angle bed = angle AEF = angle EAF
So angle EAF = angle H (equivalent substitution)
So AC = HC
Be = HC
So AC = be
Draw the auxiliary line by yourself
When you see a midline, you have to think about extending it
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 △ 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 △ 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 △ ABC, ∠ ACB = 90 °, AC = BC, be ⊥ CE at point E. ad ⊥ CE at point D Verification: △ BEC ≌ △ CDA
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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．
RELATED INFORMATIONS
- 1. 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
- 2. 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,
- 3. 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?
- 4. It is known that AD and AE are the angular bisector and height of triangle ABC respectively, ∠ B = 50 °, ead = 10 ° and ∠ C=____
- 5. 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)
- 6. 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
- 7. 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
- 8. 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
- 9. Let d be a point on the edge BC of △ ABC, and AB2 = ad2 + BD × DC
- 10. 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
- 11. 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）．
- 12. 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
- 13. 0
- 14. 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?
- 15. 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
- 16. 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
- 17. 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
- 18. 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)=
- 19. 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,
- 20. 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?