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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
Make ch ⊥ AB at h and ad at P, ∵ in RT ⊥ ABC, AC = CB, ∵ ACB = 90 °, cab = ∠ CBA = 45 °. ∵ HCB = 90 ° - ∵ CBA = 45 ° = ∵ BC midpoint is D, ? CD = BD. and ? ch ⊥ AB, ∵ ch = ah = BH. And ? PAH + ∠ APH = 90 °, PCF + ∠ CPF = 90 °, APH =
As shown in Figure 1, in the isosceles right triangle ABC and isosceles right triangle DBE, ∠ BDE = ∠ ACB = 90 ° and be is on the AB side. Take the midpoint F of AE and the middle of CD G. Connect GF. (1) the positional relationship between FG and DC is --- and the quantitative relationship between FG and DC is --- (2) if △ BDE is rotated anticlockwise about point B by 180 ° and other conditions remain unchanged, please complete figure 2. And judge whether the conclusion in (1) still holds? Please prove your conclusion
：（1）FG⊥CD,FG= CD．
(2) Extend the extension line of ED and AC to m, connect FC, FD and FM,
The quadrilateral bcmd is a rectangle
∴CM=BD．
And △ ABC and △ BDE are isosceles right triangles,
∴ED=BD=CM．
∵∠E=∠A=45°,
The △ AEM is an isosceles right triangle
And F is the midpoint of AE,
∴MF⊥AE,EF=MF,∠E=∠FMC=45°．
∴△EFD≌△MFC．
∴FD=FC,∠EFD=∠MFC．
And ∠ EFD + ∠ DFM = 90 °,
∴∠MFC+∠DFM=90°．
That is, △ CDF is an isosceles right triangle,
And G is the midpoint of CD,
∴FG= CD,FG⊥CD．
As shown in the figure, points B, C, D are on the same line, triangles ABC and CDE are equilateral triangles, be intersects AC with F, ad with CE with H (1) Verification: Triangle BCE congruent triangle ACD (2) verification: Triangle FHC is equilateral triangle
(1)∵BC=AC CD=EC ∠BCE=∠ACD=120°
The proof of triangle BCE ≌ triangle ACD
(2)∵AB‖EC
∴EF/FB=EC/AB
Similarly, AC / ed = ch / he
And ∵ AB = AC, EC = ed
∴EF/FB=EH/HC
∴FH‖BC
∴∠HFC=∠FCB=60°
And ∵ FCE = 60
∴∠FCE=∠HFC=∠CHF=60°
The proof that the triangle FHC is an equilateral triangle
It's not easy to do a problem in my senior year~
As shown in the figure: △ ABC and △ CDE are equilateral triangles
It is proved that ∵ ABC and ∵ ECD are equilateral triangles,
∴AC=BC，EC=DC，∠ACB=∠ECD=60°，
In △ BCE and △ ACD,
BC＝AC
∠ECD＝∠ACB
EC＝DC ，
∴△BCE≌△ACD（SAS），
The corresponding sides of an congruent triangle are equal
As shown in the figure, both triangle ABC and triangle CDE are equilateral triangles. AD and be intersect at point M. join MC and prove that angle BMC = angle DMC
It is proved that: over point C is C, CG ? ad is ad in G, CH ? be is in H ? equilateral ? ABC, equilateral ? CDE ? AC = BC, DC = EC, ACB ∵ DCE = 60 ? ACD ? ACD ? be in H ? be in H ? be in H ? be in h 8 be in H ? h ? h ? BCE ∵ CG ⊥ ad, CH ⊥ be
It is known that, as shown in the figure, ad is the middle line of triangle ABC. Cut CE and BF on AD and its extension line. Do you want to judge whether triangle BDF and triangle CDF are congruent? What is the relationship between BF and CE?
Triangle BDF is congruent with triangle CDE, and BF and CE are congruent: BF / / CE
It is proved that in triangle BDF and triangle CDE ∵ ad is the center line of triangle ABC
∴BD=DC ①
(2) DFF = (3) EDF
The △ BDF ≌ △ CDE (edge, angle, edge) is obtained from ①, ② and ③
Thus ∠ BFD = ∠ CED (the corresponding angles of congruent triangles are equal)
/ / BF / / Ce (the internal staggered angle is equal and the two lines are parallel)
As shown in the figure, in △ ABC, ad is the center line on BC, CE ⊥ ad is at e, BF ⊥ AD and the extension line of ad intersect F, which explains the reason of △ CDE ≌ △ BDF
It is proved that ∵ CD = BD, CED BFD = 90 ° and CDE = ∠ BDF
∴△CDE≌△BDF
As shown in the figure, ad is the center line on BC side of triangle ABC, BF is vertical AF, CE is vertical ad, then triangle BDF and triangle CDE must be congruent? Why?
prove:
∵BF⊥AF,CE⊥AD
∴∠BFD＝∠CED＝90
∵ ad is the center line on the side of BC
∴BD＝CD
∴∠BDF＝∠CDE
∴△BDF≌△CDE （AAS）
Is ab = AC, be perpendicular to e, CF perpendicular to F, be and CF to D, triangle BDF congruent CDE
Congruence
By AAS, the congruence of triangle Abe and triangle ACF can be explained, thus AE = AF
Using equal quantity minus equal quantity difference, it is shown that BF = CE
It can be explained by AAS that triangle BDF is equal to triangle CDE
It is known that: as shown in the figure, ad bisection ∠ BAC, de ⊥ AB, DF ⊥ AC, DB = DC, It is proved that △ ABC is an isosceles triangle
Prove that: ∵ ad bisection ∠ BAC (known),
/ / AD is the angular bisector of △ ABC vertex angle (the definition of angular bisector),
∵ de ⊥ AB, DF ⊥ AC (known),
▽ de = DF (property of angular bisector),
In RT △ BDE and RT △ CDF,
BD＝CD
BE＝CF ，
∴△BDE≌△CDF（HL）．
Ψ B = ∠ C (corresponding angles are equal),
The △ ABC is an isosceles triangle
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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?
- 4. 0
- 5. 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
- 6. 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）．
- 7. 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
- 8. 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
- 9. 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,
- 10. 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?
- 11. 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)=
- 12. 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,
- 13. 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?
- 14. 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______ .
- 15. In the triangle ABC, D is the midpoint of BC. If angle a is equal to 120 degrees and ab vector multiplied by AC vector equals negative 1, then the minimum value of Ad vector modulus is?
- 16. If the perimeter of a right triangle is 4 + 2, the root is 6, and the length of the central line on the hypotenuse is 2, then the area of the triangle is
- 17. Enumerate some algorithms under radical
- 18. As shown in the figure, in circle O, AB is the diameter of circle O, CD is a chord, and CD is perpendicular to point P, connecting BC and AD
- 19. In ⊙ o, the chord ab ⊥ CD is in M. the radius of ⊙ o is 5, BM = 6, am = 2. Find the size of dm-cm
- 20. Find a proof question: AB is the diameter of circle O, CD is a chord in the circle, AE ⊥ CD is in E, BF ⊥ CD is in F, and CE = DF