As shown in the figure, ad is the angular bisector of △ ABC, de ⊥ AB, DF ⊥ AC, and the perpendicular feet are e and f respectively. If AB = 5, AC = 3, and the area of △ ABC is 16, find the length of de?

De is equal to 4, and the area of triangle is equal to the sum of abd and ADC of triangle, i.e. 1 / 2Ab * de + 1 / 2Ac * DF = 16. Ad is an angular bisector. Because the distance from the point on the bisector to both sides of the corner is equal, de = DF. From ab = 5 and AC = 3, de = 4 can be obtained

As shown in the figure, points a, B, D and E are on the same straight line, ad = EB, BC ‖ DF, ∠ C = ∠ F. verification: AC = EF

Proof: ad = EB
/ / ad-bd = eb-bd, that is, ab = ed
And ∵ BC ∥ DF,
∴∠CBD=∠FDB
∴∠ABC=∠EDF
In △ ABC and △ EDF,
A kind of
∠C＝∠F
∠ABC＝∠EDF
AB＝ED
∴△ABC≌△EDF，
∴AC=EF

As shown in the figure, in △ ABC, D is a point on AC, e is a point on CB extension line, and AC BC＝EF DF， Confirmation: ad = EB

It is proved that DH ∥ BC is made through point D and ab is crossed with H, as shown in Fig,
∵DH∥BC，
∴△AHD∽△ABC，
∴AD
AC=DH
BC is ad
DH=AC
BC，
∵DH∥BE，
∴△BEF∽△HDF，
∴BE
HD=EF
DF，
And AC
BC＝EF
DF，
∴BE
HD=AD
DH，
∴AD=EB．

As shown in the figure, in △ ABC, D is a point on AC, e is a point on CB extension line, and AC BC＝EF DF， Confirmation: ad = EB

0

Why didn't you even have a
Not equal to
Because according to the meaning of the title
Let you ask for BFD and CFE congruence
But angle c = angle B
Angle FBD = 180 angle ABC
So it's impossible to be all inclusive

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, e is a point on AB, f is a point on AC extension line, and be = CF, if EF and BC intersect at D, prove: de = DF

It is proved that FH ∥ AB cross BC extension line to H,
∵FH∥AB，
∴∠FHC=∠B．
And ∵ AB = AC,
∴∠B=∠ACB．
And ∵ ACB = ∵ FCH,
∴∠FHE=∠FCH．
∴CF=HF．
And ∵ be = CF,
∴HF=BE．
And ∵ FH ∥ ab,
∴∠BED=∠HFD，
And △ in hedbe,
∠B＝∠FHC
BE＝HF
∠BED＝∠HFD ，
∴△DBE≌△FHE（ASA）．
∴DE=DF．

As shown in the figure, in △ ABC, D is the midpoint of BC, e and F are points on AB and AC respectively, and de ⊥ DF. It is proved that be + CF > EF

is this one?
Extend ed, make DG = De, connect CG and FG
Easy to get triangle DEB is equal to triangle GCD
So be = CG
Because de = DG, DF = DF, angle EFD = angle FDG = 90 degrees
So FG = EF
Because CF + DG > FG
GF=BE,FG=EF
So be + CF > EF

As shown in the figure, △ ABC, ab = AC, e is a point on AB, f is a point on AC extension line, and be = CF, if EF and BC intersect at D, prove: de = DF

As shown in the figure, ad is the angular bisector of △ ABC, de ⊥ AB, DF ⊥ AC, and the perpendicular feet are e and f respectively. If AB = 5, AC = 3, and the area of △ ABC is 16, find the length of de?