It is known that the midpoint of BC is e and the midpoint of ad is f in the regular tetrahedron ABCD. The relationship between AE and CF is judged by AE and CF 1 It is known that in the regular tetrahedron ABCD, the midpoint of BC is e, the midpoint of ad is f, even AE and CF 1. Judge the relationship between AE and CF The cosine of the angle of 2 AE, CF

It is known that the midpoint of BC is e and the midpoint of ad is f in the regular tetrahedron ABCD. The relationship between AE and CF is judged by AE and CF 1 It is known that in the regular tetrahedron ABCD, the midpoint of BC is e, the midpoint of ad is f, even AE and CF 1. Judge the relationship between AE and CF The cosine of the angle of 2 AE, CF

As shown in the figure, let the edge length of the regular tetrahedron ABCD be a,
Connect ed, take the midpoint m of ED, connect cm and FM, then FM ‖ AE, and FM = 1 / 2ae,
The angle formed by a straight line AE and CF in a different plane is called CFM or its complementary angle,
∵AE=CF=(√3/2）a,
∴FM=(√3/4)a,
In RT △ MEC, EC = (1 / 2) a, EM = (√ 3 / 4) a,
∴MC=(√7/4)a
∴cos∠CFM=
(CF² +FM² −MC² )
/(2CF•FM)＝2/3
∴∠CFM=arccos2/3

In a tetrahedron ABCD, e is the midpoint of BC, f is on edge ad, and AF: FD = 2:1. The cosine value of the angle between AE and CF is obtained

S (x) denotes the root X and 〈 denotes the angle
Connect BF, take midpoint P, remove midpoint Q of BD, connect AP, PQ, AQ and EP
In the triangle Apq, AQ = s (3) / 2, PQ = FD / 2 = 1 / 6, AQP = 30 degrees
AP = s (19) / 6 from cosine theorem
In ape, AE = s (3) / 2, PE = CF / 2 = s (7) / 6
AP=S（19）/6
Using the anti cosine theorem to get the cosine value of AEP
COS（〈AEP）=5S（21）/42
It is the cosine value of the angle between AE and CF

Known: as shown in the figure, in the square ABCD, AE ⊥ BF, the perpendicular foot is p, AE and CD intersect at point E, BF and ad intersect at point F, verification: AE = BF

It is proved that the ∵ quadrilateral ABCD is a square, AE ⊥ BF, ∵ DAE + ≌ AED = 90 °, ≌ DAE + ≌ AFB = 90 °, ∵ AED = ≌ AFB, and ∵ ad = AB, ≌ bad = ≌ D, ≌ AED ≌ ABF, ≌ AE = BF

Known: as shown in the figure, in the square ABCD, AE ⊥ BF, the perpendicular foot is p, AE and CD intersect at point E, BF and ad intersect at point F, verification: AE = BF

It is proved that the ∵ quadrilateral ABCD is a square, AE ⊥ BF, ∵ DAE + ≌ AED = 90 °, ≌ DAE + ≌ AFB = 90 °, ∵ AED = ≌ AFB, and ∵ ad = AB, ≌ bad = ≌ D, ≌ AED ≌ ABF, ≌ AE = BF