As shown in figures 1 and 2, it is known that quadrilateral ABCD is a square with a moving point p on ray AC, where PE⊥AD (or extension line) is in E, PF⊥DC (or extension line) is in F, and ray BP intersects EF in G. 。
(1) In figure 1, let the side length of square ABCD be 2, the area of quadrilateral ABFE be y, and AP=x, and find the functional expression of y about x;
(2) Conclusion: GB⊥EF is valid for both figure 1 and figure 2. Please choose any number to prove it.
(3) Please prove according to Figure 2: △ FGC ∽△ PFB.
answer
(1) solution: ∵PE⊥AD, PF⊥DC,
∴ Quadrilateral EPFD is a rectangle,
AP = x，
∴AE=EP=DF=
2
2
x，
DE=PF=FC=2-
2
2
x，
∴S quadrilateral ABFE=4-
1
2
Ed. DF-
1
2
BC? Football club
=4-
1
2
×
2
2
x(2-
2
2
x)-
1
2
×2×(2-
2
2
x)
=
1
four
x2+2；
(2) Proof: As shown in figure 1, extend FP from AB to H,
∵PF⊥DC,PE⊥AD，
∴PF⊥PE,PH⊥HB，
Namely ∠ BHP = 90,
∵ quadrilateral ABCD is a square,
∴AC split ∠DAB,
∴ Available PF=FC=HB, EP=PH,
At △FPE and △BHP,
PF=BH
∠FPE =∠ BHP Billiton
PE=HP
∴△FPE≌△BHP(SAS)，
∴∠PFE=∠PBH，
And ? ≈FPG =∠BPH,
∴△FPG∽△BPH，
∴∠FGP=∠BHP=90，
Namely GB ⊥ ef;
(3) Proof: As shown in Figure 2, connect PD, ∵GB⊥EF,
∴∠BPF=∠CFG①，
In △DPC and △BPC
DC = BC
∠DCP=∠BCP= 135
PC=PC
∴△DPC≌△BPC(SAS)，
∴PD=PB，
PD=EF，∴EF=PB，
And ∵GB⊥EF,
∴PF2=FG？ EF，
∴PF2=FG？ PB，
And PF=FC,
∴PF？ FC=FG？ PB，
∴
Pulse Frequency (abbreviation of pulse frequency)
lead
=
Floated Gyro
Football club
②,
∴△ FGC ∴△ PFB is obtained from ① and ②.
analyse
(1) According to the meaning of the question, the S quadrilateral ABFE=4-
1
2
ED×DF-
1
2
BC×FC and then get the answer;
(2) Firstly, by using the property of square, it is proved that △ FPE △ BHP (SAS) can be obtained, and then △FPG∽△BPH can be obtained.
(3) First get △ DPC △ BPC (SAS), and then get △ FGC ∽△ PFB through similar triangles's judgment.
This topic mainly examines the nature of the square, congruent triangles's judgment and similar triangles's judgment and nature. It is the key to solve the problem to skillfully apply the properties of the square and get the relationship between the corresponding angle and the corresponding edge.