As shown in Figure 1, in △ ABC, points e.d.f ab.bc.ca If ad bisects ∠ BAC, then the quadrilateral AEDF is a diamond?

As shown in Figure 1, in △ ABC, points e.d.f ab.bc.ca If ad bisects ∠ BAC, then the quadrilateral AEDF is a diamond?

Because AD and EF share equally
So quadrilateral AEDF is parallelogram
In △ AED and △ AFD
Because ad bisects ∠ BAC
Therefore, ead = fad
And ∠ AED = ∠ AFD
AD=AD
△AED≌△AFD(AAS)
AE=AF
The quadrilateral AEDF is a diamond

In the triangle ABC, AB is equal to AC, and the point DEF is the midpoint of the three sides of the triangle ABC respectively
It's urgent

Proof: because D, e and F are the middle points of three sides of triangle ABC respectively
So de.ef is the median line of triangle ABC
So de = 1 / 2Ac
AD=BD=1/2AB
AF=CF=1/2AC
EF=1/2AB
Because AB = AC
So ad = de = EF = AF
So the quadrilateral ADEF is a diamond

As shown in the figure, C and D are two points on the line ab. if CB = 4cm, DB = 6cm, and D is the midpoint of AC, how many centimeters is the length of AC equal to?
It's a process!

DB-CB=CD=6-4=2
D is the midpoint of AC
AC=CDx2=2x2=4cm

As shown in the figure, C and D are two points on line ab. if CB = 4cm, D and B = 7cm, and D is the midpoint of AC, the length of line AC is calculated

It should be DB = 7 > CB = 4, so the order of the four points is a, D, C, B
DC=BD-CB=7-4=3cm
Because D is the midpoint of AC, AC = 2CD = 3 * 2 = 6cm

As shown in the figure, ad ‖ BC, EF passes through the midpoint o of AC. (1) verification: OE = of;
(2) If the line EF rotates around point O and intersects with AD and BC at points e 'and f', is there still OE '= of'?
(3) Where is the shortest line segment cut by AD and BC when EF rotates around point o?

It can be proved by making the vertical lines of AD and BC from point o
Question 2 is valid
In 3, the line segment of vertical time section is the shortest

Given that point AB = CD, ad = BC, O is the midpoint of AC, the straight line passing through point O intersects ad, BC at point EF respectively, which shows the principle of OE = of

AB=CD,AD=BC
A quadrilateral ABCD is a parallelogram
O is the midpoint of AC, Ao = Co
So, in △ OAE and △ COF
Because, FCO = OAE, AOE = COF
△AOE≌△COF
So, OE = of

As shown in the figure, ab = CD, ad = BC, O is the midpoint of AC, OE ⊥ AB is in E, of ⊥ D is in F

∵AB=CD,AD=BC
The quadrilateral ABCD is a parallelogram
∴∠BAC=∠DCA
That is, EAO = FCO
∵ o is the midpoint of AC
∴OA=OC
⊙ OE ⊥ AB vs e, of ⊥ CD
∴∠OEA=∠OFC=90°
∴△AOE≌△COF(AAS)
∴OE=OF

As shown in the figure, given that point D is on BC, BD: DC = 2:1, point E is on ad, AE: ed = 2:3, the extension line of be intersects AC at point F, find the value of be: EF

Let DH ∥ AC intersect BF at h, BH: HF = BD: DC = 2:1 = 10:5, ∥ DHE ∥ AFE. ∥ EF: eh = AE: ed = 2:3, ∥ BH: HF = 10:5. ∥ be: EF = (BH + he): EF = 13:2

As shown in the figure, ab = AC, DB = DC, f is a point on the extension line of AD

It is proved that: in △ abd and △ ACD, ab = acbd = CDAD = ad ≌ abd ≌ ACD, ≌ bad = CAD, in △ BAF and △ CAF, ab = AC ≌ BAF = CAF = AF ≌ BAF ≌ CaF (SAS), ≌ BF = CF

As shown in the figure, ad is the middle line of △ ABC, f is the point on AC, CF = 2AF, connecting BF to ad at point E. verification: be = 3ef

It is proved that: crossing point D as DG ∥ BF, crossing AC to g, DH ∥ AC, crossing BF to point h, ∥ DH ∥ AC, DG ∥ BF, ∥ quadrilateral HDGF is parallelogram, ∥ HD = FG, DG = HF, ∥ ad is the middle line of △ ABC, ∥ DB = DC = 12bc, DH = 12fc, ∥ DH ∥ FC, D is the middle point of BC, ∥ BH = HF, ∥ DG ∥ BF, ∥ fgfc =