Mathematical angle bisector of the first volume of the second day of junior high school

If it passes through C and the vertical line AB passes through F, there will be △ CEB △ CFD.

This is because: 1, CE=CF (angular bisector property)

2. Angle CEB= Angle CFD=90 degrees

3, EB=DF (because

①AE=AF，②AD+AB=2AE

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Since there is no map, I assume that AB> is in the year 200.

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AB=AE+EB

AD=AF-DF

AE=AF

So AE+EB+AF-DF=2AE.

Simplify and own

EB=DF)

Therefore ∠CBE=∠CDF.

So the two angles mentioned in the title complement each other.

① ②→③Right.

Let the intersection of AD and EF be H.

AE=AF from the nature of the bisector, that is, △AEF is isosceles △

therefore

Angle dividing line

Vertical bottom edge, proof

① ③→②Wrong.

② ③→ ① Right.

Remember that the midpoint of AD is m, connecting me and MF.

Use me = MF = 0.5ad.

MH is

Male marginal

, < mhe = < mhf = 90 degrees

△MHF?△MHE

Therefore, HF=HE.

AE=AF，

Properties of angle dividing line

So △ AEH △ AFH

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