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
Connect AE
Because it's Square and am = FN
So am / AC = an / AF
It is concluded that Mn is parallel to FC
FC belongs to planar CBE
Mn ‖ plane CBE
RELATED INFORMATIONS
- 1. 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
- 2. ABCD and abef are parallelograms, m and N are the points on diagonal AC and BF respectively, and am: FN = AC: BF
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. The quadrilateral ABCD is a rectangle,
- 9. As shown in the figure, the quadrilateral ABCD is a rectangle, ∠ EDC = ∠ cab, ∠ Dec = 90 ° (1) verification: AC ‖ de( 2) Make BF ⊥ AC at point F through point B, connect EF, try to judge the shape of quadrilateral BCEF, please explain the reason (do not judge parallelogram)
- 10. As shown in the figure, the quadrilateral ABCD is a rectangle, ∠ EDC = ∠ cab, ∠ Dec = 90 ° (1) Verification: AC ‖ de; (2) Make BF ⊥ AC at point F through point B, connect EF, try to judge the shape of quadrilateral BCEF, and explain the reason
- 11. As shown in the figure, the planes of two congruent squares ABCD and abef intersect AB, m ∈ AC, n ∈ FB, and am = FN
- 12. 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
- 13. In parallelogram ABCD, de ⊥ AC in E, BF ⊥ AC in F, connect be ` DF, then be ∥ DF? Be = DF? Try to explain your reason
- 14. 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
- 15. 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
- 16. The quadrilateral ABCD is a parallelogram, de ⊥ AC, BF ⊥ AC, and the perpendicular feet are e, F, be and DF respectively
- 17. 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!
- 18. In the parallelogram ABCD, points E and F are points on AB and CD respectively, de / / BF
- 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. As shown in the figure, parallelogram ABCD, e, f are two points on AC, de ‖ BF, verification: AE = CF