As shown in the figure, two pieces of paper of equal width are crossed and overlapped, and the quadrilateral ABCD of the overlapped part is______ Shape

As shown in the figure, two pieces of paper of equal width are crossed and overlapped, and the quadrilateral ABCD of the overlapped part is______ Shape

Through point a, make AE ⊥ BC in E, AF ⊥ CD in F, ∵ the width of two strips is the same, ∵ AE = AF. ∵ ab ∥ CD, ad ∥ BC, ∵ quadrilateral ABCD is parallelogram. ∵ s ▱ ABCD = BC · AE = CD · AF. also ∵ AE = AF. ∵ BC = CD, ∵ quadrilateral ABCD is diamond

Two pieces of paper of equal width are crossed and overlapped. Is the overlapping part ABCD a diamond?
Why?
Tell me the reason
Come on! Wait on the line

Yes
∵AB‖CD
The quadrilateral ABCD is a parallelogram
∵ s quadrilateral ABCD = CD * H1 (one strip width)
=BC * H2 (another note width)
And ∵ H1 = H2
∴CD=BC
The quadrilateral ABCD is a diamond

As shown in the figure, two pieces of paper of equal width are crossed and overlapped, and the quadrilateral ABCD of the overlapped part is______ Shape

Through point a, make AE ⊥ BC in E, AF ⊥ CD in F, ∵ the width of two strips is the same, ∵ AE = AF. ∵ ab ∥ CD, ad ∥ BC, ∵ quadrilateral ABCD is parallelogram. ∵ s ▱ ABCD = BC · AE = CD · AF. also ∵ AE = AF. ∵ BC = CD, ∵ quadrilateral ABCD is diamond

As shown in the figure, is the quadrilateral ABCD formed by the superposition of two rectangles of equal width a diamond? Why?

The quadrilateral ABCD is a diamond
A quadrilateral ABCD is a parallelogram
Through a vertex to the other two sides of the vertical line, equal width and height equal, get congruent, get edge equal
Without a picture, I can only describe it in words

As shown in the figure, it is known that f is a point on the edge ab of diamond ABCD, DF intersects AC at e. the angle AFD = angle CBE is proved

Certification:
Connect BD
∵ the diamond diagonals are perpendicular to each other
∴ED=EB
∴∠EDB=∠EBD
∵ diamond diagonal bisects diagonal
∴∠ABD=∠CBD
∵∠AFD=∠ABD+∠EDB
∠CBE=∠CBD+∠EBD
∴∠AFD=∠CBE

It is known that, as shown in the figure, the quadrilateral ABCD is a diamond, f is a point on AB, DF intersects AC with E

It is proved that: the ∵ quadrilateral ABCD is a diamond, BC = CD, ab ∥ CD, and ∥ AFD = CDE. In △ BCE and △ DCE, BC = CD, BCE = dcece = CE ≌ BCE ≌ DCE, CBE = CDE, AFD = CDE, AFD = CBE

It is known that, as shown in the figure, the quadrilateral ABCD is a diamond, f is a point on AB, DF intersects AC with E

It is proved that: the ∵ quadrilateral ABCD is a diamond, BC = CD, ab ∥ CD, and ∥ AFD = CDE. In △ BCE and △ DCE, BC = CD, BCE = dcece = CE ≌ BCE ≌ DCE, CBE = CDE, AFD = CDE, AFD = CBE

It is known that, as shown in the figure, the quadrilateral ABCD is a diamond, f is a point on AB, DF intersects AC with E

It is proved that: the ∵ quadrilateral ABCD is a diamond, BC = CD, ab ∥ CD, and ∥ AFD = CDE. In △ BCE and △ DCE, BC = CD, BCE = dcece = CE ≌ BCE ≌ DCE, CBE = CDE, AFD = CDE, AFD = CBE

As shown in the figure, in diamond ABCD, ∠ bad = 80 °, the intersection of the vertical bisector of AB, the diagonal AC at point F, e is the perpendicular foot, connecting DF, then ∠ CDF is equal to ()
A. 60°B. 65°C. 70°D. 80°

Connect BD, BF, ∫ bad = 80 ° and ∫ ADC = 100 ° and ∫ EF vertical bisection AB and AC vertical bisection BD, ∫ AF = BF, BF = DF, ∫ AF = DF, ∫ fad = ∫ FDA = 40 ° and ∫ CDF = 100 ° - 40 ° = 60 °. So select a

As shown in the figure, known quadrilateral ABCD is diamond, de ⊥ AB, DF ⊥ BC, please explain the quantitative relationship between be and BF

EB = BF. In △ ade and △ CDF, ∵ quadrilateral ABCD is rhombic, ∵ a = ≌ C, ad = CD. De ⊥ AB, DF ⊥ BC, ∵ AED = ≌ CFD = 90 ·. ≌ ade ≌ CDF. ≌ AE = cf. EB is equal to BF