In the triangle ABC, ad is the center line on BC side, e is a point on AC, be and ad intersect F, if AE = EF, it is proved that BF = AC Here I know to extend ad, but I don't know if you can make CG through point C, hand over ad to g, make CG = DC, I don't know if this is OK. Ask!

Guo Dun Yong replied: ∵ AE = EF,  EAF = ∠ EFA, and ∠ EFA = ∠ BFD (antiparietal angle), make the extension line of BK ∥ AC crossing ad at K,

As shown in the figure, in △ ABC, ad is the center line on BC edge, e is a point on AC, be and ad intersect at point F. if AE = EF, it is proved that AC = BF

Extend ad to m to make ad = DM. Connect BM
Delta ADC congruent △ BDM
∠DAC=∠M
AE=EF
∠DAC=AFE=∠∠BFD
∠M=∠BFD
BF=BM=AC

As shown in the figure, ad is the middle line of △ ABC, be intersects AC to e, and ad to F, and AE = EF. Verification: AC = BF

It is proved that ∵ ad is the midline of ᙽ ABC,
∴BD=CD．
Method 1: extend ad to point m, make MD = FD, connect MC,
In △ BDF and △ CDM,
BD＝CD
∠BDF＝∠CDM
DF＝DM
∴△BDF≌△CDM（SAS）．
∴MC=BF，∠M=∠BFM．
∵EA=EF，
∴∠EAF=∠EFA，
∵∠AFE=∠BFM，
∴∠M=∠MAC，
∴AC=MC，
∴BF=AC；
Method 2: extend ad to point m, make DM = ad, connect BM,
In △ ADC and △ MDB,
BD＝CD
∠BDM＝∠CDA
DM＝DA ，
∴△ADC≌△MDB（SAS），
∴∠M=∠MAC，BM=AC，
∵EA=EF，
Ψ cam = ∠ AFE, and ∠ AFE = ∠ BFM,
∴∠M=∠BFM，
∴BM=BF，
∴BF=AC．

As shown in the figure, in △ ABC, ad is the midline on BC, e is a point of AC, be and ad intersect at point F, if AE = EF. Verification: AC = BF

It is proved that: extend ad to g, make DG = ad, connect BG, in △ BDG and △ CDA, ? BD = CD ≌ BDG ≌ △ CDA (SAS),  BG = AC, ≌≌△ CDA (SAS), ? BG = AC, ≌≌△ CDA (SAS), ? AE = ef  CAD = ∠ AFE ∠ BFG ≌ ≌ △ CDA (SAS), ? CAD = ≌ ≌ ≌ ≌ ≌ ≌ ≌ ≌ ≌ ≌

As shown in the figure, in diamond ABCD, e is the midpoint of AD, and the extension line of EF ⊥ AC crossing CB is at F Verification: AB and EF are equally divided

Proof: connect BD, AF, be,
In diamond ABCD, AC ⊥ BD
∵EF⊥AC，
/ / EF ∥ BD, ed ∥ FB,
The quadrilateral EDBF is a parallelogram, de = BF,
∵ e is the midpoint of AD,
∴AE=ED，∴AE=BF，
AE ‖ BF,
The quadrilateral aebf is a parallelogram,
That is, AB and EF are equally divided

As shown in the figure, in diamond ABCD, e is the midpoint of AD, and the extension line of EF ⊥ AC crossing CB is at F Verification: AB and EF are equally divided

As shown in the figure, in diamond ABCD, e is the midpoint of AD, and the extension line of EF ⊥ AC crossing CB is at point F (1) Is de equal to BF? Please state the reasons (2) Connecting AF and be, is the quadrilateral afbe parallelogram? Give reasons

(1) In diamond ABCD, BD ⊥ AC, ∵ EF ⊥ AC, ∵ eg ⊥ BD, ∵ e is the mid point of AD, ᙽ eg is the median line of △ abd, ∵ Ag = BG, and ∵ ad ∥ BC,

It is known that in △ ABC, BD and CE are two high lines, f is a point on BD, G is a point on CE extension line, BF = AC, CG = ab ① Please judge the shape of △ AFG ② When f is a point on BD reverse extension line and G is a point on CE reverse extension line, other conditions remain unchanged, whether the conclusion in ① still holds? Please draw a figure and prove your conclusion ② Please draw the picture in

The shape of △ AFG is isosceles right triangle
In △ CEA, ∠ ACE + ∠ CAE = 90 degrees; in △ BDA, ∠ abd + ∠ bad = 90 degrees,
So ∠ ace = ∠ abd
In △ GCA and △ ABF, AC = BF, GC = AB, so △ GCA ≌ △ ABF, so Ag = AF
And ∠ ace = ∠ AGC + ∠ GAC = ∠ GAC + ∠ BAF
Since ACE + EAC = 90 degrees, AGC + GAC = 90 degrees, so △ fag is an isosceles right triangle

In the known triangle ABC, CE is perpendicular to AB and BF is perpendicular to ac. it is proved that AEF of triangle is similar to ACB of triangle

It is shown that the triangle ABF is similar to the triangle ace, and AE: AC = AF: AB is obtained. Because the angle a is common, the triangle AEF is similar to the triangle ACB

In the triangle ABC, ad is the center line on BC side, e is a point on AC, be and ad intersect F, if AE = EF, it is proved that BF = AC Here I know to extend ad, but I don't know if you can make CG through point C, hand over ad to g, make CG = DC, I don't know if this is OK. Ask!