As shown in the figure, AB is parallel to CD, ad and BC intersect at point O, EF passes through point O and intersects AB, CD at e and f respectively, and AE = DF. It is proved that O is the midpoint of EF

As shown in the figure, AB is parallel to CD, ad and BC intersect at point O, EF passes through point O and intersects AB, CD at e and f respectively, and AE = DF. It is proved that O is the midpoint of EF

Proof: because AB is parallel to CD
So angle a = angle D
Because AE = DF
So AE is parallel to DF
So the angle AEF = the angle DFE
Because {angle a = angle D AE = DF angle AEO = angle DFO
So triangle AEO = triangle DFO (ASA)
So EO = fo
So o is the midpoint of EF

As shown in the figure, the chords AC and BD in circle O intersect with F, EF is made through point F, parallel to AB, the extension line crossing CD is at e, and the tangent line eg of circle O is made through point E, and G is the tangent point. It is proved that EF = eg If the picture, you should be able to draw it by yourself. Sorry, I can't draw it

According to the tangent theorem, we can know that eg 2 = ed * EC, let EF and circle intersect Mn, (n point on AC arc) ∠ ECB = 1 / 2 (BM radians + MD radians) ∠ EFD = 1 / 2 (an radians + MD radians) because AB is parallel to EF and Mn is on EF, Mn is parallel to AB, and the arc length of an is equal to that of BM

As shown in the figure, △ ABC, ab = AC, de parallel Ba crossing AC to e, DF parallel CA crossing AB to F, connecting EF.AD Do you have the following conclusions? Explain the reasons 1. AD and EF are equally divided 2.AE=BF. Draw your own picture. There's no time!

1 de parallel AB DF parallel AC
Then afde is a parallelogram, so its diagonal ad EF is equally divided
2 de parallel ab
Angle EDC = ABC = ACD
Therefore, EC = ed = AF
AE=AC-EC=AB-AF=BF

In △ ABC, ab = AC, point D is the midpoint of BC, ∠ BDF = ∠ CDE, the extension lines of De and Ba intersect at point E, and the extension lines of DF and Ca intersect at point F, connecting EF Verify EF ‖ BC

Because AB = AC and D is the midpoint of BC, so ∠ bad = ∠ cadad is perpendicular to BC to D, because ∠ BDF = ∠ CDE, then ∠ FDA = ∠ EDC, because ∠ bad = ∠ CAD (proved) ad = ad, then △ FDA is equal to △ EDF, then △ AFE is isosceles △ and the height of EF side passes D point (AD extension line is perpendicular to EF)

As shown in the figure, ad is the angular bisector of the triangle ABC, de / / Ca intersects AB at e, DF / / BA intersects AC at F

Proof: because Da bisection angle BAC
So: angle ead = angle fad
De|ac, df|ab
So: angle EDA = angle fad; angle FDA = angle EAD
That is: angle ead = angle EDA; angle FDA = angle fad
Therefore: EA = EB; FA = FD
AEDF is a parallelogram, so it can be concluded that AEDF is a diamond
Ad is perpendicular to ef

In △ ABC, the midpoint F of AB is de ⊥ BC, the perpendicular foot is point E, and the extension line of intersection CA is at point D. if EF = 3, be = 4, ﹤ C = 45 °, find the length of DF

Because De is perpendicular to BC, the angle Dec = 90 ° because the angle c = 45 ° so the angle d = 45 ° so de = EC,
Because f is the midpoint of AB, FG is the median line of the triangle ABC, so EF = eg = 3, because be = 4, so BG = 7, so GC = 7, because ed = EC, EF = eg, so DF = GC = 7

As shown in the figure, in △ ABC, AC = BC, D is a point on Ca, e is a point on CB extension line, and AD= BE.DE Let's Cross AB to point F to prove DF = EF

It is proved that the crossing point D is DG ∥ BC and crosses AB with G
∵AC＝BC
∴∠A＝∠ABC
∵DG∥BC
∴∠AGD＝∠ABC,∠GDF＝∠E,∠DGF＝∠EBF
∴∠A＝∠AGD
∴AD＝DG
∵AD＝BE
∴DG＝BE
∴△DGF≌△EBF （ASA）
∴DF＝EF
The math group answered your question,

As shown in the figure, it is known that ad is the chord of circle O, D is the midpoint of arc BC, De is the tangent of circle O and intersects with the extension line of chord AB at point E= AC.AE

Arc BD = arc CD ∠ bad = ∠ CAD
That is ∠ DAE = ∠ CAD
De is circle O tangent ∠ EDB = ∠ BCD ∠ BDA = ∠ BCA
∠EDB+∠BDA=∠BCD+∠BCA
∠EDA=∠DCA
△AED∽△ADC
AE:AD=AD:AC
AD^2=AC*AE

It is known that: as shown in the figure, BD is the diameter of ⊙ o, BC is the chord, a is the midpoint of BC arc, AF ∥ BC intersects the extension line of dB at point F, ad intersects BC at point E, AE = 2, ed = 4 (1) It is proved that AF is tangent of ⊙ o; (2) Find the length of ab

(1) Proof: connect OA,
∵ A is the midpoint of BC arc,
∴OA⊥BC．
∵AF∥BC，
∴OA⊥AF．
⊙ AF is the tangent of ⊙ o
（2）∵∠BAE=DAB，∠ABE=∠ADB，
∴△ABE∽△ADB．
∴AB
AD=AE
AB．
∴AB2=AE•AD=12．
∴AB=2
3．

It is known that: as shown in the figure, BD is the diagonal of the parallelogram ABCD, O is the midpoint of BD, EF ⊥ BD is at point O, and it intersects with AD and BC at points E and f respectively. Verification: de = DF

It is proved that in the parallelogram ABCD, ad ∥ BC,
∴∠OBF=∠ODE
∵ o is the midpoint of BD
∴OB=OD
In △ BOF and △ doe,
A kind of
∠OBF＝∠ODE
OB＝OD
∠BOF＝∠DOE
∴△BOF≌△DOE
∴OF=OE
∵ EF ⊥ BD at point o
∴DE=DF．