As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD

It is proved that in RT △ AEC, AF ⊥ EC,
∴AC2=CF•CE．
In RT △ ABC, ad ⊥ BC,
∴AC2=CD•CB．
∴CF•CE=CD•CB．
∴CF
CB＝ CD
CE．
∵∠DCF=∠ECB，
∴△DCF∽△ECB．
∴∠B=∠CFD．

As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD

As shown in Figure 1, in △ ABC, D, e and F are the midpoint of the three sides, G point is on the edge AB, the perimeter of △ BDG and quadrilateral acdg is equal, let BC = a, AC = B, ab = C (1) Find the length of segment BG; (2) Verification: DG bisection ∠ EDF; (3) Connect CG, as shown in Figure 2. If △ BDG is similar to △ DFG, verify: BG ⊥ CG

(1) ∵ △ BDG has the same perimeter as the quadrilateral acdg,
∴BD+BG+DG=AC+CD+DG+AG，
∵ D is the midpoint of BC,
∴BD=CD，
∴BG=AC+AG，
∵BG+（AC+AG）=AB+AC，
∴BG=1
2（AB+AC）=1
2（b+c）；
(2) It is proved that ∵ points D and F are the midpoint of BC and ab respectively,
∴DF=1
2AC=1
2b，BF=1
2AB=1
2c，
And ∵ FG = bg-bf = 1
2（b+c）-1
2c=1
2b，
∴DF=FG，
∴∠FDG=∠FGD，
∵ points D and E are the midpoint of BC and AC respectively,
∴DE∥AB，
∴∠EDG=∠FGD，
∴∠FDG=∠EDG，
That is, DG bisection ∠ EDF;
(3) It is proved that ∵ △ BDG is similar to △ DFG, ∵ DFG > b, ∠ bgd = ∠ DGF (common angle),
∴∠B=∠FDG，
From (2), it is concluded that: ∠ FGD = ∠ FDG,
∴∠FGD=∠B，
∴DG=BD，
∵BD=CD，
∴DG=BD=CD，
The points B, G and C are on the circumference with BC as the diameter,
∴∠BGC=90°，
BG ⊥ CG

As shown in the figure, in △ ABC, D is the midpoint of BC, the straight line GF passing through point d intersects AC with F, parallel line BG crossing AC with point G, de ⊥ GF, crossing AB at point E, connecting eg (1) Results: BG = CF; (2) Please judge the relationship between be + CF and EF and prove your conclusion

It is proved: (1) that: (1)

As shown in the figure, in △ ABC, ∠ BAC = 90 °, BG bisection ∠ ABC, GF ⊥ BC at point F, ad ⊥ BC at point D, intersection BG at point E, and connect EF (1) The results showed that: ① AE = Ag; ② quadrilateral aefg was rhomboid (2) If ad = 8, BD = 6, find the length of AE

It is proved that: (1) ① ∵ BG bisection  ABC,  Abe = ∵ DBE,  Abe + ∠ age = 90 °, EBD + ∠ DEB = 90 °, GEA = ∠ bed,  AEG = ∠ EGA, i.e. Ag = AE. ② ? GF ⊥ BC at point F, ad ⊥ BC at point D, BG bisection  ABC, ? CFG = ∠ CDA = 90 °, ad ∥ GF, Ag = GF, and ?

As shown in the figure, in the triangle ABC, the bisector of angle B and the bisector of angle c intersect at D, DG parallel to BC, AC, AB at F, G. verification: GF = BG minus CF

Because dg / / BC
So ∠ GDB = ∠ DBC
Because BD is an angular divider
Therefore, abd = DBC, so GDB = GBD, BG = GD
Because CD is an angular divider dg / / BC
So ∠ GDC = ∠ FCD
So CF = DF
Because GF = gd-fd
So GF = bg-cf

As shown in the figure, in △ ABC, ∠ BAC = 90 °, BG bisection ∠ ABC, GF ⊥ BC at point F, ad ⊥ BC at point D, intersection BG at point E, and connect EF (1) The results showed that: ① AE = Ag; ② quadrilateral aefg was rhomboid (2) If ad = 8, BD = 6, find the length of AE

As shown in the figure, △ ABC, ab = AC, points D and E are on the extension line of AB and AC respectively, and BD = CE, de and BC intersect at point F. verification: DF = EF

It is proved that DG ∥ AE is made through point D and passed to point BC at point G, as shown in the figure,
∴∠1=∠2，∠4=∠3，
∵AB=AC，
∴∠B=∠2，
∴∠B=∠1，
∴DB=DG，
And BD = CE,
∴DG=CE，
In △ DFG and △ EFC
∠4＝∠3
∠DFG＝∠EFC
DG＝CE ，
∴△DFG≌△EFC，
∴DF=EF．

As shown in the figure, △ ABC, ab = AC, points D and E are on the extension line of AB and AC respectively, and BD = CE, de and BC intersect at point F. verification: DF = EF

As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at D, e is a point on AB, AF ⊥ CE is at F, ad intersects CE at G point. It is proved that ∠ B = ∠ CFD