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In rectangular ABCD, point m is the midpoint of CD, ab = 8cm, BC = 10cm. Fold the rectangular ABCD so that a and M coincide. Crease EF intersects ad at point E, BC at point F, Find the length of AE
The key is to know that the crease is the vertical bisector of am
Assuming that am intersects EF with O, it is easy to prove that △ AOE is similar to △ Adm
AE/AO=AM/AD
Because Ao = (1 / 2) am, AE = (1 / 2) am ^ 2 / ad = (1 / 2) * (AD ^ 2 + DM ^ 2) / ad = (1 / 2) * (10 ^ 2 + 4 ^ 2) / 10 = 5.8
As shown in the figure, fold the rectangle with length ad = 10 cm and width AB = 8 cm along AE, so that point d falls on point F on BC side, and calculate the length of de
In RT △ ABF, BF = 6cm, FC = bc-bf = 10-6 = 4cm; in RT △ EFC, FC2 + CE2 = ef2, (8-de) 2 + 42 = ef2, EF = de = 5cm
As shown in the figure, square ABCD and square abef intersect AB, m and N are points on BD and AE respectively, and an = DM
It is proved that Mn ‖ plane EBC
As shown in the figure, make mg ∥ Da & nbsp; ∵ Da ∥ CB & nbsp; ∥ mg ∥ CB, & nbsp;
∵AN＝DM, AE＝DB＝√2AB.
∴AN/AE＝DM/DB＝AG/GB, ∴NG‖RB
∵MG‖CB, NG‖RB
Ψ planar GMN ‖ planar EBC. & nbsp; & nbsp;
∵ Mn ∈ plane GMN
Ψ Mn ‖ plane EBC
It is known that two congruent rectangles ABCD and abef with common edge AB are different in a plane, P and Q are points on diagonal AE and BD respectively, and AP = DQ. The BCE of PQ / / plane is proved
Because of congruent rectangles ABCD and abef, BD = AE
Because AP = DQ, there is a point m on AB such that DQ: BQ = am: BM = AP: EP
So there are MQ / / BC, MP / / be
So face QMP / / face BCE
So PQ / / BCE
As shown in the figure, in rectangular ABCD, ab = 4, BC = 5, AF bisects ∠ DAE, EF ⊥ AE, then CF equals ()
A. 23B. 1C. 32D. 2
∵ quadrilateral ABCD is a rectangle, ∵ ad = BC = 5, ∵ d = ∵ B = ∵ C = 90 °, ∵ AF bisection ∵ DAE, EF ⊥ AE, ∵ DF = EF, according to Pythagorean theorem: AE2 = af2-ef2, ad2 = af2-df2, ∵ AE = ad = 5, in △ Abe, according to Pythagorean theorem: be = AE2 − AB2 = 3, ∵ EC = 5-3 = 2, ∵ BAE + ∠ AEB = 90 °
As shown in the figure, in rectangular ABCD, ab = 4, BC = 5, points E and F are on BC and CD respectively, connecting AE, AF, EF and AF, bisecting ∠ DAE
AE ⊥ EF to find the value of CF length and Tan ∠ ECF
ABCD is a rectangle
∴AB=CD=4,AD=BC=5
∴∠D=∠B=∠C=90°
∵AE⊥EF
∴∠AEF=∠D=90°
∵ AF bisects ∠ DAE, i.e. ∠ fad = ∠ FAE
AF=AF
∴△AFD≌△AFE(AAS)
∴AE=AD=5
EF=DF
In RT △ Abe: be = √ (AE & # 178; - AB & # 178;) = √ (5 & # 178; - 4 & # 178;) = 3
∴EC=BC-BE=5-3=2
DF=EF=CD-CF=4-FC
In RT △ CEF: EF & # 178; = CF & # 178; + EC & # 178;
(4-CF)²=CF²+2²
CF=12/8=3/2
2. Tan ∠ ECF is meaningless, ∠ ECF = 90 degree
As shown in the figure, in ladder ABCD, ad ‖ BC, point E is on BC, AE = be, and AF ⊥ AB, connect EF. (1) if EF ⊥ AF, AF = 4, ab = 6, find the length of & nbsp; AE. (2) if point F is the midpoint of CD, prove: CE = be-ad
(1) Let em ⊥ AB intersect AB at points M. ∵ AE = be, EM ⊥ AB, ∵ am = BM = 12 × 6 = 3; ∵ EF ⊥ AF, ∵ ame = {MAF =} AFE = 90 °, ∵ quadrilateral amef is rectangle, ∵ EF = am = 3; in RT △ AFE, AE = af2 + ef2 = 5; (2) extend the intersection of AF and BC at points n. ∵ ad ∥ en, ∵ DAF =} n
As shown in the figure, in rectangular ABCD, EF is the vertical bisector of BD. given BD = 20, EF = 15, calculate the perimeter of rectangular ABCD
Let length AB = x, width BC = y, ∵ - dab = 90 ° = - EOB = 90 °, ∵ - B = - B, ∵ RT △ DAB ∵ RT △ EOB, ∵ ABOB = bdbe, ∵ ad = BC, DF = be, ∵ x2 + y2 = 202xy = 2152 = 43, (5 points) the solution is X1 = 16y1 = 12, X2 = - 16y2 = - 12 (rounding off), (7 points) ∵ the circumference of the rectangle is 56. (8 points)
Rectangle ABCD, EF is the vertical bisector of BD, BD = 20, EF = 15?
According to the angle, tanadb = 3 / 4, and then according to the Pythagorean theorem, the length and width are respectively 12 and 9, and the area is 108
As shown in the figure, in rectangular ABCD, EF is the point on AB and CD respectively, and AF = CE. Try to judge whether AF and CE are parallel and explain the reason
D---F---C
| / /|
| / / |
| / / |
A--E----B
Not in the middle
∵ AE = CF and ABCD is rectangular
∴EB=DF AD=BC
∴Rt△EBC≌Rt△FDA
∴∠CEB=∠DFA
∵AB‖DC
∴∠FAE=∠DFA
∴∠FAE=∠CEB
∴AF‖CE
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