In parallelogram ABCD, de ⊥ AC in E, BF ⊥ AC in F, connect be ` DF, then be ∥ DF? Be = DF? Try to explain your reason

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• 1. In the parallelogram ABCD, EF is the midpoint of AD and BC respectively The intersection of the center line BD and CE of triangle ABC with O, F and G is the midpoint of OB and OC respectively
• 2. As shown in the figure, the planes of two congruent squares ABCD and abef intersect AB, m ∈ AC, n ∈ FB, and am = FN
• 3. It is known that ABCD and abef are two squares and not in the same plane, m and N are the points on the diagonal AC and FB respectively, and am = FN It is suggested that the extension line connecting an and be should be connected to g
• 4. As shown in the figure, let ABCD and abef be parallelograms, they are not in the same plane, m and N are the points on the diagonal AC and BF respectively, and am: FN = AC: BF (Continued), prove: Mn ‖ plane bec
• 5. ABCD and abef are parallelograms, m and N are the points on diagonal AC and BF respectively, and am: FN = AC: BF
• 6. Parallelogram ABCD parallelogram abef is on the same side AB, m, n are on the diagonal AC, BF respectively, and am: AC = FN: FB to prove Mn / / plane ADF
• 7. As shown in the figure, the diagonal lines AC and BD of rectangular ABCD intersect at point O, EF ⊥ BD at point O, ad at point E, BC at point F, and EF = BF
• 8. As shown in the figure, in square ABCD, the diagonal lines AC and BD intersect at point O, and the points EF are on AC and BD respectively, and BF = CE connects be, AF.AF There is a relation between be and quantity
• 9. 1. As shown in the figure, the quadrilateral ABCD is a rectangle, ∠ EDC = ∠ cab, ∠ Dec = 90 ° (1) prove: AC ‖ de; (2) make BF ⊥ AC at point F through point B, connect
• 10. ABCD is a rectangle ∠ EDC = ∠ cab ∠ Dec = 90 ° (1) verification: AC ‖ de (2) through B as BF ⊥ AC connected to f to judge the shape of becf and explain the reason
• 11. In a parallelogram, points E and F are on AB and CD respectively, DF = be. Is the quadrilateral debf a parallelogram? Tell me your reason
• 12. As shown in the figure, in the angle ABCD, the points E and F are on AB and CD respectively, be = DF. It is proved that the quadrilateral debf is a parallelogram The graph is inside, and the line DB is removed
• 13. The quadrilateral ABCD is a parallelogram, de ⊥ AC, BF ⊥ AC, and the perpendicular feet are e, F, be and DF respectively
• 14. E. F is the two points on the diagonal AC of the parallelogram ABCD, de ⊥ AC, BF ⊥ AC, proof: the parallelogram debf is a parallelogram, urgent now!
• 15. In the parallelogram ABCD, points E and F are points on AB and CD respectively, de / / BF
• 16. 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
• 17. As shown in the figure, parallelogram ABCD, e, f are two points on AC, de ‖ BF, verification: 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. In the parallelogram ABCD, the bisector BC of ∠ bad intersects at point E and the bisector DC intersects at point F (1) In Figure 1, it is proved that CE = CF; (2) If ∠ ABC = 90 ° and G is the midpoint of EF (as shown in Figure 2), write the degree of ∠ BDG directly; (3) If ∠ ABC = 120 ° FG ‖ CE, FG = CE, connect dB and DG respectively (as shown in Figure 3), and calculate the degree of ∠ BDG
• 20. In the parallelogram ABCD, point E is on the edge AB, and point F is on the edge BC. The areas of △ ade, △ bef, and △ CDF are 5, 3, and 4 respectively. The area of △ CDE is calculated This is an exercise in the chapter "quadrilateral" in the eighth grade mathematics volume II of pep. Since you can't draw pictures here, please draw your own pictures according to the meaning of the title. It won't be difficult to draw Sorry, the problem should be to find the area of △ def.