In the triangle ABC of the right graph, ad: DC = 2:3, AE = EB

In the triangle ABC of the right graph, ad: DC = 2:3, AE = EB

AD:DC=2:3
Then ad: AC = 2:5
If AE = EB, then the height h 1 on the ABC side of the triangle and the height H 2 on the side ad of the triangle AED have H1 = 2h2
be
Area ratio of a to triangle ABC = (2:5) * (1:2) = 1:5
Then the area ratio of a and B is 1:4

As shown in the figure, △ ABC, AE intersects BC at point D, ∠ C = ∠ e, ad = 4, BC = 8, BD: DC = 5:3, then the length of De is equal to () A. 20 Three B. 15 Four C. 16 Three D. 17 Four

∵∠ADC=∠BDE，∠C=∠E，
∴△ADC∽△BDE，
∴AD
BD＝DC
DE，
∵AD=4，BC=8，BD：DC=5：3，
∴BD=5，DC=3，
∴DE=BD•DC
AD=15
4．
Therefore, B

As shown in the figure, CD bisects the extension line of ∠ ACB, AE ‖ DC to BC at point E, if ∠ ace = 80 °, then ∠ CAE=______ Degree

∵∠ACE=80°，
∴∠ACB=100°，
And ∵ CD bisection ∵ ACB,
∴∠DCA=100°×1
2=50°，
∵AE∥DC，
∴∠CAE=∠DCA=50°．

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 that the point of intersection of B, C, B, C is greater than that of B, C, B, C, B, C, B, C, B, C, B, C, B, C, B, C, B, C, B, C, B, C, B, C, B, C, B, C, B, C

The ∠ BOD is a complementary angle of the triangle AOB, ∠ BOD = ∠ Bao + ∠ ABO ad, Bo, Co are the bisectors of ∠ BAC, ∠ ABC and ∠ ACB respectively, so ∠ BOD = 1 / 2 ∠ BAC + 1 / 2 ∠ ABC = 1 / 2 (∠ BAC + ∠ ABC) = 1 / 2 (180 - ∠ ACB) = 90-1 / 2 ∠ ACB (the sum of inner angles is equal to 180) in EOC of right triangle, ∠ EOC

In △ ABC, it is known that ab = 3, AC = 6, BC = 7, ad is the ∠ BAC bisector. It is proved that DC = 2bd

Certification: as shown in the figure,
Pass through point D as de ∥ AB to AC at point E
Then ∠ EDA = ∠ DAB, and ∠ DAB = ∠ EAD,
∴∠EDA=∠EAD，
∴EA=ED．
∵DE∥AB，
∴CD
DB＝CE
EA＝CE
ED＝AC
AB=6
3=2，
∴CD=2DB．

In the triangle ABC, ad is perpendicular to BC, D. f is the point above ad, and DF = DC, connecting BF and extending AC to point E 1. If ad = BD, please explain the reason why be is vertical AC 2. If be is perpendicular to AC, is ad = BD? Please explain the reason This question comes from the eighth question on page 16 of mathematics class training (curriculum standard Zhejiang Education Edition)

1、BD=AD,DF=CD
RtΔADC≌RtΔBDF
∠DAC=∠DBF,∠ACD=∠BCE
ΔBCE∽RtΔBDF
Δ BCE is RT Δ
BE⊥AC
2. No
AD=BD,∠DBA=∠DAB
Be ⊥ AC, it can only be proved that ⊥ DBE = ∠ DAC
As long as DF ≠ DC
You can't get ad = BD. there are countless triangles

As shown in the figure, in △ ABC, ab = ad, DC = BD, de ⊥ BC, de intersects AC at point E, be intersects ad at point F The results showed that: (1) BDF △ CBA; (2) AF = DF

It is proved that: (1) BD = DC, de ⊥ BC,  EB = EC.  EBD = ∵ C. (3 points)

As shown in the figure, in triangle ABC, AD bisects angle BAC, BE is perpendicular to point E, AB is perpendicular to point F, DF is perpendicular to point AC, BD=DC, to verify: angle B=

, ∵ ad bisection ∠ BAC,
DE⊥AB,DF⊥AC,
BD=DC
ν AB = AC (three lines in one isosceles triangle),
DE=DF,
∠BED=∠CFD=90°
∴∠B=∠C

As shown in Figure 1, in triangle ABC, ad is perpendicular to BC at point D, and ad = BD, point F is on ad, and BF intersects AC at point E. if DF = DC, what is the position of BF and AC What is the quantitative relationship? Are you sure?

∵AD⊥BC
∴∠BDF=∠ADC=90°
∵BD=AD DF=DC
∴△ACD≌△BDF
∴BF=AC ∠BFD=∠ACD
∵∠BFD=∠AFE
∴∠AFE=∠ACD
∵∠DAC=∠FAE
∴△ACD∽△AEF
∴∠AEF=∠ADC=90°
∴BE(BF⊥AC
ν BF = AC and BF ⊥ AC