In square ABCD, point G starts from point C and moves along CB; point h starts from point a and moves along Ba, and two points move at the same speed and stop at the same time

In square ABCD, point G starts from point C and moves along CB; point h starts from point a and moves along Ba, and two points move at the same speed and stop at the same time

(1) Making FM vertical CD from F
FM ⊥ CD, so ∠ GFM + ∠ MGF = 90
DH ⊥ FG, so ∠ EDG + ∠ MGF = 90
Therefore, GFM = edg
∠FMG=∠DCH
DC=FM
So △ FMG ≌ DCH, FG = DH
(2) Connect be
In △ BCE and △ DCE,
BC=DC,∠BCM=∠DCM=45,CM=CM
So △ BCE ≌ △ DCE, ∠ EBC = ∠ EDC
CD ⊥ BC, therefore, EDC + DHC = 90
DH ⊥ PF, so ∠ Eph + ∠ DHC = 90. ∠ EDC = ∠ Eph
Therefore, EBC = Eph, EB = EP
Because ∠ EBC + ∠ EBF = 90, and ∠ EBC + ∠ EBF + ∠ PFB + ∠ Eph = 180
Therefore, PFB = EBF, EB = EF
So EP = EF
De is the middle line on FP in △ DFP and the high line on FP
So △ DFP is an isosceles triangle, de = DP

If AB = 6cm, ad = 4cm, then EF = cm, the parallelogram ABCD, AF, be bisects ∠ DAB, ∠ CBA, intersection CD and points F, e respectively
In a parallelogram ABCD, an acute angle c is the height CE and CF of the sides AB and ad. if the angle between the two heights is 135 degrees, then the internal angles of the parallelogram are degrees

EF=2CM
45 degrees 135 degrees 45 degrees 135 degrees

As shown in the figure, in the parallelogram, ∠ ABC = 75 °. AF ⊥ BC is in F, AF intersects BD in E, if de = 2Ab, ∠ AED=______ °．

ABCD is a parallelquadrilateral, and its \\\\\\\\abcdis a parallelogram, \ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ ° and ∠ ADO= Therefore, the answer is 65 degrees

In the parallelogram ABCD, through point B, make be, perpendicular to CD and connect AE, f is the point above AE, angle BFE = angle c, try to explain the triangle ABF and triangle EAD

In the parallelogram ABCD, ∠ C + ∠ ADC = 180 °, ∠ BFE + ∠ BFA = 180 °, ∫ BFE = ∠ C
The results show that the values of BFA = ∠ ade, ∵ ab ‖ BC, ∵ AED = ∠ EAB
The Δ ABF is similar to the triangular EAD

In the quadrilateral ABCD, ad ∥ BC, e are the midpoint of CD, linking AE, be, be ⊥ AE, extending AE and extending line point F of BC
Verification: (1) AE = Fe
（2）AB=BF

(1) Because of AD / / BC, so the angle DAE = angle EFC, and because e is the midpoint, so de = EC, so ade and FCE are congruent, AE = Fe
(2) Because triangles are congruent, AE = EF, e is the midpoint of AF, and because be is perpendicular to AF, the triangle ABF is isosceles right triangle, so AB = BF

As shown in the figure: in the quadrilateral ABCD, ad is parallel to BC, ab = AD + BC, e is the midpoint of CD, to prove AE ⊥ be

Extend ad to F, make DF = BC, then AF = ab,
∵AD∥BC
The BCFD is a parallelogram
∵ e is the midpoint of CD
The AE is the vertical bisector of BF
∴AE⊥BE

(2008 · Guang'an) as shown in the figure, in ladder ABCD, ad ‖ BC and E are the midpoint of CD, connecting AE and extending the extension line between AE and BC at point F
(1) (2) if ad = 2, ab = 8, when BC is, point B is on the vertical bisector of line AF, why?

(1) It is proved that: ∵ ad ∥ BC, ∵ f = ∠ DAE. (1 point) and ∵ FEC = ∠ AED, ∵ ECF = ∠ ade, ∵ e is the midpoint of CD, ∵ CE = de. in △ FEC and △ AED,