As shown in the figure, it is known that AB is parallel to CD, OA = OD, BC passes through point O, e, f are on the straight line AOD and AE = DF

As shown in the figure, it is known that AB is parallel to CD, OA = OD, BC passes through point O, e, f are on the straight line AOD and AE = DF

Proof: because ab ∥ CD
So ∠ Bao = ∠ CDO
Because AOB = ∠ cod, Ao = OD
So Δ AOB ≌ Δ cod
So Bo = do
Because Ao = do, AE = FD
So EO = fo
Because EO = fo, ∠ EOB = ∠ COF, Bo = do
So Δ EOB ≌ Δ COF
So ∠ EBO = ∠ FCO
So EB is parallel to CF

Ad ⊥ BF, AE ⊥ CF, the vertical feet are D and e respectively, and DF = EF, ∠ BAF = ∠ CAF, verification, ∠ B = ∠ C

Do geometry problems must draw pictures, students
I haven't been a junior high school student for a long time. I'll just tell you the process. The expression may not meet the current teaching requirements. You can change it yourself
prove:
Right triangle AFD and AFE
Because AF = AF, DF = EF
So they are congruent triangles
So ∠ DAF = ∠ EAF
Because ∠ BAF = ∠ CAF, ∠ DAF = ∠ EAF
So ∠ bad = ∠ CAE
In right triangle bad and CAE
Because ∠ bad = ∠ CAE, another acute angle ∠ B = ∠ C

It is known that EF ∥ BC, AE? 2 = ad · ab. verification: DF ∥ EC A D E F B C A. D, e, B are collinear, a, F, C are collinear, connecting DF, EF, EC, BC

EF ‖ BC, parallel line divided into straight line, proportional
AE/AB=AF/AC,
AE²=AD·AB,AD/AE=AE/AB=AF/AC
DF‖EC

In trapezoidal ABCD, point E is on AB, and ad = a, BC = B. as shown in the figure, if AE: EB = DF: FC = m: n, judge whether EF and BC are parallel, please prove your conclusion

EF ∥ BC proves that if f is a parallel line of AB, then it intersects BC at point h, and the extension line intersecting ad is at point G, then the quadrilateral abhg is a parallelogram  AB = GH, easy proof ∨ FDG ∨ FCH ᦻ GF / FH = dg / CG = AE / be = m / N ? GF / GH = AE / AB ∵ AB = GH ? AE = AE = GF ? aefg is a parallelogram  EF ∥ Ag ∥ BC (2) connects BD

As shown in the figure, in the isosceles trapezoid ABCD, if ad ∥ BC, diagonal AC ⊥ BD are at point O, AE ⊥ BC, DF ⊥ BC, the vertical feet are e, F, ad = 4, BC = 8, then AE + EF is equal to______ .

Extend BC to G so that DG ∥ AC, ∵ ad ∥ BC, ∵ adgc is a parallelogram, ? DG = AC, ∵ AC ⊥ BD, ∵ DG ⊥ BD, and ∵ isosceles trapezoid ABCD, ? AC = BD, ∵ DG = BD, ? DBG is an isosceles right triangle,  BG = 2bd2 ∵ BD = DG = 62 ∵ DF ⊥ BG, ? DBG is an isosceles right triangle,   

AB is the diameter of circle O, AC and BD are tangent lines of circle O, CD tangent circle O is at e, EF is perpendicular to AB, AB, AD are intersected at points E and g respectively to prove eg = FG

It should be: EF is perpendicular to AB, AB, ad at points F and g. it is proved that ∵ AC and BD, CD are tangent lines of circle O ᙽ CA = CE, BD = ed and AC ⊥ AB, BD ⊥ AB, EF ⊥ ab ⊥ AC / / EF / / DB

The two chords AB and Cd in the circle intersect at e, passing through e as EF, parallel BC crossing ad extension line at F, making tangent FG of circle through F, proving: FG = Fe

prove:
∵EF//BC
∴∠DEF=∠DCB
∵ DAB = ∠ DCB
∴∠DEF=∠DAB
And ? DFE = ∠ EFA [common angle]
∴⊿DFE∽⊿EFA（AA’）
∴FE/FD=FA/FE
It is transformed into Fe? 2 = FD × FA
∵ FG is the tangent of a circle
 = FD × Fa [cutting line theorem]
∴FE²=FG²
∴FE=FG

In the triangle ABC, ab = AC, D is the moving point on AB, passing through point D as DF vertical BC, crossing BC at point F, crossing CA extension line at E 1. Try to judge the relationship between AD and AE and explain the reasons 2. If point D is on the extension line of Ba, and other conditions remain unchanged, please draw a figure and answer directly whether the conclusion in (1) still holds

(1) Ad = AE. The reasons are as follows:
Because AB = AC, angle B = angle C. angle BAE = angle B + angle c = 2 angle B
Because DF is perpendicular to BC, angle BFD = 90 degrees, angle B + angle BDF = 90 degrees
Because angle ade = angle BDF, angle ade = 90 degrees - angle B
Because the angle e + angle ade + angle BAE = 180 degrees, the angle e + 90 degrees - angle B + 2 angles B = 180 degrees,
So, angle e = 90 - angle B = angle ade, so ad = AE
(2) When point D is on the extension line of Ba and E is on line AC, other conditions remain unchanged
The reasoning method is the same

In any triangle ABC, D, e, f are the midpoint of AB, BC and ac. it is proved that ab ∥ EF, BC ∥ DF, CA ∥ de are proved

Extend DF to G so that FG = DF and connect CG
∵AF=CF,∠AFD=∠CFG,DF=GF,
∴△ADF≌△CGF,
∴CG=AD,∠A=∠ACG,
∴CG=BD,CG∥BD,
The parallelogram BCGD,
∴DF∥BC
Similarly, ab ∥ EF, CA ∥ de can be proved

It is known that ad is an angular bisector of △ ABC, where de ‖ AC intersects AB at point E and DF ‖ AB intersects AC at point F Verification: quadrilateral AEDF is rhombic

Proof: ∵ DE ∵ AC, DF ∵ AB,
The quadrilateral AEDF is a parallelogram,
∵ ad is the angular bisector of ∵ ABC,
∴∠1=∠2，
∵DE∥AC，
∴∠2=∠3，
∴∠1=∠3，
∴AE=DE，
The quadrilateral AEDF is a diamond (a parallelogram with equal adjacent sides is a diamond)