1, triangle ABC, angle A = 60, and the intersection of the bisector BE of ∠B and ∠C and CD and point o: OE=OD.

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.