As shown in the figure, in square ABCD, points E and F are on AB and BC respectively, and AE = BF, AF and de intersect at point g. from the given conditions, what conclusions can you draw? Why?

As shown in the figure, in square ABCD, points E and F are on AB and BC respectively, and AE = BF, AF and de intersect at point g. from the given conditions, what conclusions can you draw? Why?

Conclusion: AF = De
AF ⊥ De is g
Because ABCD is a square, so ad = ab ∠ ABF = ∠ DAE = 90 ° and AE = BF, so triangle ABF is equal to triangle DAE, so AF = de ∠ BAF = ∠ ade ∠ AFB = ∠ DEA ∠ BAF + ∠ AFB = 90 °
Then there is ∠ AGD = ∠ BAF + ∠ DEA = ∠ ade + ∠ DEA = 90 degree
So AF ⊥ de

In square ABCD, e is the point on BC, BF ⊥ AE intersects CD with F, then AE: AF

AE: I don't know
But AE: BF = 1:1
Because △ Abe is equal to △ BFC

In square ABCD, e and F are points on AB and BC respectively, and AE = BF

This is very easy
Proband △ Abf ≌ △ DAE
∵∠AFB=∠DEA
And ∵ ∠ AFB + ∠ Fab = 90
∴∠DEA+∠FAB=90
∴FA⊥DE

Through the four vertices of the quadrilateral AB, CD, make the parallel lines of diagonal AC and BDD respectively, the resulting figure is a diamond, then
If the figure obtained by crossing the four vertices of the quadrilateral AB CD and making the parallel lines of diagonal AC and BD respectively is a diamond, then the original quadrilateral must be; ()
A parallelogram
B quadrilateral with equal diagonal
C rectangle
A quadrilateral whose diagonals are perpendicular to each other
Explain why!

Choose a
No matter what the figure is, it is very easy to prove that the figure composed of diagonal and those parallel lines is four parallelogram lines, and each side of the figure is the diagonal of the four parallelograms. Therefore, it can be proved that it is a parallelogram by the equality of this angle and that angle
This is a train of thought. Do it yourself