In the parallelogram ABCD, CE ⊥ CD is passed through point C, ad is crossed with point E, and the segment EC is rotated 90 degrees counterclockwise around point e to obtain the segment EF

In the parallelogram ABCD, CE ⊥ CD is passed through point C, ad is crossed with point E, and the segment EC is rotated 90 degrees counterclockwise around point e to obtain the segment EF

In &; ABCD, pass through point C to make CE ⊥ CD, intersect ad with point E, and rotate segment EC 90 ° counterclockwise around point e to get segment EF, as shown in the figure
(1) Research on drawing in the picture:
① When P1 is any point on the extension line of the line segment CD, connect. EP1, rotate the line segment EP1 90 ° counterclockwise around the point e to get the line segment EG1, judge the position relationship between the line FG1 and the line CD, and explain the reason; (draw in Figure 1)
② When P2 is any point on the extension line of line segment DC, connect EP2 and rotate EP2 90 ° counterclockwise around point e to get line segment EG2. Judge the position relationship between line fg2 and line CD, draw a figure and write your conclusion directly. (draw in Figure 2)
(2) Under the condition of (1), connect FP1 and p1g1, if EP1 = 8, ad = 6, AE = 1, AB: CE = 3:4, calculate the area of △ p1g1f
(1) (1) the position relationship between line FG1 and line CD is perpendicular to each other
Proof: as shown in Figure 1, let the intersection of line FG1 and line CD be H
∵ segments EC and EP1 rotate 90 ° counterclockwise around point e respectively to obtain segments EF and EG1,
∴∠P1EG1=∠CEF=90°,EG1=EP1,EF=EC．
∵∠G1EF=90°-∠P1EF,∠P1EC=90°-∠P1EF,
∴∠G1EF=∠P1EC．
∴△G1EF≌△P1EC．
∴∠G1FE=∠P1CE．
∵EC⊥CD,
∴∠P1CE=90°,
The value of g1fe is 90 degrees
The value of EFH is 90 degrees
And FHC = 90 degrees
∴FG1⊥CD．
② According to the requirements of the title, the drawing is shown in Figure 1. The position relationship between line g1g2 and line CD is perpendicular to each other
(2) A quadrilateral ABCD is a parallelogram
∴∠B=∠ADC
∵AD=6,AE=1,AB：CE=3：4,
∴DE=5,CD：CE=3：4
CE = 4
From (1), we can conclude that the quadrilateral fech is a square
∴CH=CE=4
(1) as shown in Figure 2, P1 is on the extension line of segment ch,
∵FG1=CP1
∴S△P1FG1=×FG1•P1H
In RT △ ecp1, EP1 = 8
CP1=FG1=4
∴P1H=4-4
∴S△P1FG1=×=24-8．

In the quadrilateral ABCD, AB is parallel to CD, ad is parallel to BC, e and F are on AD and CD respectively, and CE is equal to AF, CE and AF intersect at point P

If BF is connected, the area of △ ABF = 1 / 2 area of parallelogram ABCD
If be is connected, the area of △ BCE = 1 / 2 area of parallelogram ABCD
The area of Δ ABF = △ BCE
∵AF=CE
The heights on AF and CE are equal, that is, the distances from point B to AF and CE are equal
So B is on the bisector of APC
∴∠APB=∠BPC
So Pb bisects APC

If AB = a, ad = B, CE = m, find the length of BF
In △ ABC, C = 90 °, BC = 8 cm, AC ∶ AB = 3 ∶ 5, point P starts from point B, moves along BC to point C at a speed of 2 cm / s, and point Q starts from point C, moves along CA to point a at a speed of 1 cm / s. if P and Q start from B and C at the same time, after how many seconds, the triangle with C, P and Q as vertices is exactly similar to △ ABC
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1. It is proved that the triangle CEF is similar to the triangle DEA, and then Ce: de = CF: Da, that is, (A-M): M = B: CF, so CF = b * [M: (A-M)] = (BM): (A-M), BF = BC + CF = B + [BM: (A-M)] = AB: a-m2. First, AC: ab: BC = 3:5:4, because BC = 8cm, AC = 6cm, ab = 10cm