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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
(1) Delta DEF is an equilateral triangle
The proof is as follows:
∵ △ ABC is an equilateral triangle,
∴∠A=∠B=∠C，AB=BC=CA，
And ∵ ad = be = CF,
ν DB = EC = FA, (2 points)
≌ △ ADF ≌ △ bed ≌ △ CFE, (3 points)
/ / DF = de = EF, that is, △ DEF is an equilateral triangle; (4 points)
(2) Ad = be = CF holds
The proof is as follows:
As shown in the figure, ∵ △ DEF is an equilateral triangle,
∴DE=EF=FD，∠FDE=∠DEF=∠EFD=60°，
∴∠1+∠2=120°，
ABC is an equilateral triangle,
∴∠A=∠B=∠C=60°，
∴∠2+∠3=120°，
Ψ 1 = ∠ 3, (6 points)
Similarly, ∠ 3 = ∠ 4,
≌ △ ADF ≌ △ bed ≌ △ CFE, (7 points)
/ / ad = be = cf. (8 points)
In the acute triangle ABC, ad, be and CF are the heights on the three sides respectively. It is proved that the vertical center h of the triangle ABC is the heart of the triangle def
∵ ahe = ∠ Bhd, AC perpendicular to be, ad perpendicular to BC
∴∠CAD=∠EBC
∴sin∠CAD=sin∠EBC
∴CE/BC=CD/AC
∵ in △ CDE and △ cab
∠ECD=∠BCA
∴△CDE∽△CAB
∴∠CDE=∠CAB
In the same way, we can get ∠ BDF = ∠ cab
∴∠CDE=∠BDF
∴∠ADF=∠ADE
In the same way, we can get ∠ bef = ∠ bed,
∠CFD=∠CFE
/ / AD, be, CF are the three angular bisectors of △ def
The vertical center h of △ ABC is the heart of △ def
As shown in the figure, be and CF are the two heights of △ ABC. G and H are the midpoint of EF and BC respectively. What is the relationship between GH and ef? Explain the reasons
GH⊥EF
Proof: link EH and FH
Be ⊥ EC, CF ⊥ FB. ⊥ BEC and ⊥ CFB are right triangles. ∵ EH and FH are the midlines on the hypotenuse of the two triangles respectively. ∵ eh = FH = 1 / 2BC
In the isosceles △ EFH, Hg is the center line of the bottom. According to the three lines in one isosceles triangle, Hg is also the height on the bottom
As shown in the figure, the triangle ABC is an equilateral triangle with a side length a, and P is any point in the triangle ABC. Through the point P, EF ∥ AB intersects AC, BC at points E and F, GH ∥ BC intersects AB, AC in G, h, and Mn ∥ AC intersects AB, BC in M and N. please guess what the value of EF + GH + nm is? Whether the value changes with the position of point P, and explain the reason for your conjecture I want a reason, because so don't copy the good work online
∵ △ ABC is an equilateral triangle,
∴∠A＝∠B＝∠C＝60°．
∵GH‖BC,∴∠AGH＝∠B＝60°,∠AHG＝∠C=60°．
﹥ AGH is an equilateral triangle, ﹥ GH = Ag = am + Mg ①
Similarly, △ BMN is an equilateral triangle, ν Mn = MB = Mg + GB
∵MN‖AC,EF‖AB,
The quadrilateral AMPF is a parallelogram, PE = am
In the same way, it can be proved that the quadrilateral bfpg is a parallelogram, ﹥ pf = GB
∴EF=PE+PF＝AM+GB． ③
① + 2 + 3
EF+GH+MN=AM+GB+MG+GB+AM+MG=2(AM+MG+GB)=2AB=2a．
22. (10 points) as shown in the figure, the straight line EF ∥ GH, points B and a are respectively on the lines EF and GH, connecting AB, and making a triangle ABC on the left side of ab, Where ∠ ACB = 90 ° and ∠ DAB = ∠ BAC, the straight line BD bisects ∠ FBC, and the intersection line GH is at D (1) If point C is exactly on EF, as shown in Figure 1, then ∠ DBA=________ (2 points) (2) If point a is moved to the left and other conditions remain unchanged, as shown in Figure 2, does the conclusion in (1) still hold? If so, prove your conclusion; if not, explain your reason. (6 points) (3) If the title condition "∠ ACB = 90 °" is changed to "∠ ACB = 120 °", other conditions remain unchanged ∠DBA= _________ (write the result directly without proof) (2 points)
(1) The process of 60 degree is the same as (2)
As shown in the figure, △ ABC is the inscribed triangle of ⊙ o, the diameter GH ⊥ AB, and the extension lines of AC to D, GH, BC intersect at E (1) It was proved that: ∠ oad = ∠ E; (2) If od = 1, de = 3, try to find the radius of ⊙ o; (3) When What type of arc is AGB, the outer center of △ CED is on the outside, inside and one side of △ CED
0
It is known that in △ ABC, ad is the bisector of ∠ BAC. It is proved that BD: DC = AB: AC
It is proved that, as shown in the figure, the parallel line of ad passing through C intersects the extension line of BA at point E,
∴∠DAC=∠ACE，∠BAD=∠E，
∵ ad is the bisector of ∵ BAC,
∴∠BAD=∠DAC．
∴∠ACE=∠E，
∴AC=AE，
∵CE∥AD，
∴BD：DC=BA：AE，
∴BD：DC=AB：AC．
It is known that in △ ABC, ad is the bisector of ∠ BAC. It is proved that BD: DC = AB: AC
It is proved that, as shown in the figure, the parallel line of ad passing through C intersects the extension line of BA at point E,
∴∠DAC=∠ACE，∠BAD=∠E，
∵ ad is the bisector of ∵ BAC,
∴∠BAD=∠DAC．
∴∠ACE=∠E，
∴AC=AE，
∵CE∥AD，
∴BD：DC=BA：AE，
∴BD：DC=AB：AC．
It is proved by sine theorem that if in triangle ABC, the bisector ad of angle A and the extension of side BC intersect at point D, then It is proved by sine theorem that if in triangle ABC, the bisector ad of the outer angle of angle a intersects with the extension line of side BC at point D, then BD ratio DC = AB ratio AC. note whether it is bisector of external angle
It is proved that because the bisector ad of the outer angle of angle a (angle CAE) intersects with the extension of side BC at D
So angle DAE = angle CAD
So sin angle CAD = sin angle DAE
Because angle DAE + bad = 180 degrees
So sin angle DAE = sin angle bad
So sin angle CAD = sin angle bad
In triangle CAD and triangle bad, the sine theorem is used to obtain the following results
DC / Xin angle CAD = AC / sin angle ADC
BD / Xin angle bad = AB / sin angle ADC
So DC / AC = BD / ab
So BD / DC = AB / AC
In Δ ABC, it is proved by sine theorem that AB / AC = BD / DC How to draw the picture?
Proof: to prove AB / AC = BD / DC, it can be proved that AB / BD = AC / DC
According to the sine theorem, AB / BD = sin ∠ ADC / sin ∠ bad (1)
AC/DC=sin∠ADC/sin∠CAD
CAD = ∠ 1
So: AC / DC = sin ∠ ADC / sin ∠ 1 (2)
And ∠ bad + ∠ 1 = 180 °
So, basin = 1 ∠
According to (1) (2) (3), AB / BD = AC / DC
So: AB / AC = BD / DC
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