Take the G point on BC so that BD=BG.
Because ∠ a = 60.
So ∠ BOC = 120.
Because ∠DOB=∠EOC (diagonal)
So DOB = EOC = 60 (360- 120)/2.
Especially the △ dbo △ bog of SAS.
So do = G0 ∠ DOB = ∠ GOB = 60.
So ∠ GOC = ∠ BOG = 60.
Then get △ OGC △ OEC from ASA.
So OG=OE
Because OD=OG
So OE=OD
2. It is known that when △AB=AC, ∠ A = 90, AB=AC, AE⊥BD is in E, ∠ ADB = ∠ CDF, AE is extended to BC to F, which proves that D is the midpoint of AC.
Let d be the symmetry point g of BC connecting FG and CG.
Because angle ADB= angle BAF and angle FDC= angle BAF.
Angle B= angle c = 45.
So angle AFB = 180- angle B- angle BAF = 180- angle C- angle CDF= angle DFG.
So angle AFD+ angle DFG= angle AFD+ angle DFC+ angle AFB = 180.
So lines a, f and G***
And because angle CAG= angle ABD.
Angle ACG = 2 * 45 = 90 = bad angle.
So all the bad triangles are equal to triangle ACG.
So CG=AD
CG=DC
So AD=DC
3. In the known triangle ABC, AD is the center line of BC side, E is a point above AC, and BE and AD intersect at F. If AE=EF, it is verified that AC=BF.
Expand AD to m so that DM = ad, even BM, CM.
AD=DM,BD=CD
∴ABMC is a parallelogram (diagonal bisection)
∴AC‖BM,AC=BM (equal to the last use)
∴∠DAC=∠DMB(∠DAC is ∠EAF, ∠DMB is ∝∞∠bmf used below) (the internal angles are equal) ...
In the triangle AEF,
AE = EF
∴∠EAF=∠EFA (isosceles triangle) ... ②
∫∠EFA =∠BFM (relative vertex angle is equal) ... ③
From ① ② ③, we get ∠EAF=∠EFA=∠BFM=∠BMF.
In a triangular BFM,
∠∠BFM =∠BMF
∴ Triangle BF=BM is an isosceles triangle with side length BF = BM.
AC=BM,AC = BF。
4. Given the triangle ABC, AD is the middle line on the side of BC, E is a point on AC, AD and BE intersect at point F, AE=EF. BF=AC?
Parallel lines that extend AD and pass through point b and are AC intersect at point g.
Then AC//BG, AE=EF,
Available BF=BG
In triangle BDG and triangle CDA
BD=CD,& ltADC = & ltGDB,& ltDBG = & lt; ACD,
Consistency of two triangles
So AC=BG=BF
5. In △ ABC, ∠ACB is a right angle, ∠ B = 60, AD and CE are bisectors of ∠BAC and ∠BCA respectively, and AD and CE intersect at point F, which proves that FE=FD.
Proof: FM⊥BC in M, FN⊥AB in N.
∫∠B = 60
∴∠MFN= 120
∫AD and CE are angular bisectors.
∴FM=FN
∠FAC+∠FCA= 15 +45 =60
∴∠AFC= 120
∴∠EFD= 120
∴∠EFN=∠DFM
FE = FM,∠FNE=∠FMD
∴△FEN≌△FMD
∴FD=FE
6. Point C is on BD, AC is perpendicular to BD at point C, BE is perpendicular to AD at point E, and CF = CD, so are AD and BF equal? Why?
Equality. Because AC is perpendicular to BD and BE is perpendicular to AD, triangle ACD and triangle BCF are right triangles. And because CF=CD, triangle ACD and triangle BCF are congruent (one side of each corner is equal). So AD and BF are equivalent.
7. In triangle ABC, AB=AC, and AD is high. Proof: bad angle = angle CAD.
AB=AC, AD=AD, angle ADB= angle ADC=90 degrees, so all triangles ABD are equal to triangle ACD, so angle BAD= angle CAD.
1, a rose
"Then you can take me to see my mother. But uncle, my mother lives far away from here. 」
"If I had known, I wouldn't have taken you." This gentleman said