It is known that EF in parallelogram ABCD is two points on diagonal BD, and be = DF. It is proved that bfde is parallelogram

It's impossible. It's said that EF is two points on BD and bfde is four points on a straight line

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1. It is known that, as shown in the figure, e and F are the triangles of diagonal AC of parallelogram ABCD
2. As shown in the figure, the quadrilateral ABCD is a diamond, f is the intersection of a point DF on AB, AC at e. the angle AFD = angle CBE is proved
3. As shown in the figure, in the quadrilateral ABCD, ad ∥ BC, ∠ B = 90 °, ad = 16cm, ab = 4cm, BC = 21cm, the moving point P starts from point B and moves to point C at the speed of 2cm / s along the direction of line BC, the moving point Q starts from point a and moves to point d at the speed of 1cm / s along the direction of line ad, and points P and Q start from point B and a respectively. When point P moves to point C, point Q stops moving, and the time of motion is set (1) find the length of DQ (expressed by the algebraic expression of T); (2) when the value of T is, the area of △ PQD is equal to 12cm2? (3) Is there a point P such that △ PQD is a right triangle? If it exists, request all the values of t that meet the requirements; if not, explain the reason
4. As shown in the figure, angle B = angle c = 90 degrees, M is the midpoint of BC, am bisector angle DAB, verify DM bisector angle ADC
5. It is known that: as shown in the figure, ∠ B = ∠ C = 90 °, M is the midpoint of BC, DM bisects ∠ ADC. (1) prove that am bisects ∠ bad; (2) try to explain the position relationship between DM and am? (3) What is the relationship between CD, AB and ad? Write the results directly
6. As shown in the figure, ∠ B = ∠ C = 90 °, point m is the midpoint of BC, am bisects ∠ DAB
7. As shown in the figure, in the quadrilateral ABCD, ab = ad, CB = CD, point P is the point on AC, PE ⊥ BC is in E, PF ⊥ CD is in F, proving: PE = PF
8. In ABCD, ab = AC, CB = CD, point P is a point on diagonal AC, PE is perpendicular to BC and E, PF is perpendicular to CD and F, PE = PF is proved
9. It is known that in square ABCD, P is a point on diagonal AC, PE ⊥ AB, PF ⊥ BC, and the perpendicular feet are e and f respectively
10. In square ABCD, P is the moving point on ad, PE ⊥ AC is in E, PF ⊥ BD is in F, PE + PE = 5, then the circumference of square ABCD is?
11. It is known that E and F are two points on diagonal AC of parallelogram ABCD, and be = DF. It is proved that bfde is a parallelogram
12. The bisector of the outer angle fac of a triangle is AE, angle 1 equals angle 2, ad equals AC. verify that DC is parallel to Ae
13. It is known that ab = AC, BD = DC, AE bisects ∠ fac, Q: is AE perpendicular to ad? Why?
14. It is known that in △ ABC, ab = AC, BD = DC, AE bisects ∠ fac, try to guess the position relationship between AE and ad, please explain the reason
15. In the parallelogram ABCD, e is the midpoint of AD, the extension line of be is at point F, connecting Ce (1), proving: CD = DF (2) if ad = 2CD, please write all right angle triangles in the graph In the parallelogram ABCD, e is the midpoint of AD, the extension line of be is at the point F, connecting Ce (1), proving: CD = DF (2) if ad = 2CD, please write all right triangles and isosceles triangles in the graph
16. As shown in the figure, in the parallelogram ABCD, e is the midpoint of CD, and the extension line of be and ad intersects at point F Million urgent A. still have half an hour to hand in homework a. a a... will add ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD ADD
17. It is known that, as shown in the figure, e and F are two points on the diagonal AC of the parallelogram ABCD, AE = CF
18. It is known that, as shown in the figure, e and F are two points on the diagonal AC of the parallelogram ABCD, AE = CF
19. It is known that, as shown in the figure, e and F are two points on the diagonal AC of the parallelogram ABCD, AE = CF
20. In the parallelogram ABCD, e f is two points on the diagonal AC, and AE = CF