1. Figure 2, if point P is on line A0, it does not coincide with points A and O. PE is perpendicular to PB, and PE is cross-checked with CD at point E: DF=EF.
Write the line segment PC, PA, CE.
An equivalent relationship between them and prove your conclusion.
As shown on the left.
Connect BE and PD, the intersection point p is the vertical line of AD, and the vertical foot is g.
Because point o is the AC midpoint of the diagonal of the square ABCD.
Therefore, point O is the center of the square.
And AC divides ∠DAB and ∠DCB equally.
PE⊥PB and BC⊥CE are all famous.
So, b, c, e and p are four * * * cycles.
Therefore, ∠ PEB = ∠ PCB = 45, ∠ PBE = ∠ PCE = 45.
Therefore, ∠ pbe = ∠ peb = 45.
Therefore, △PBE is an isosceles right triangle.
So, PB=PE
However, in △PAB and △PAD:
AB=AD (known)
∠ BAP =∠ DAP = 45 (certification)
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So, △ PAB △ Pad (SAS)
So, PB=PD
So: PE=PD
And PF⊥CD
So, DF=EF
Because PG⊥AD's PF⊥CD
And PCF = PAG = 45.
So △PCF and △PAG are isosceles right triangles.
Furthermore, the quadrilateral DFPG is a rectangle.
So:
PA=√2*PG
PC=√2*CF
And PG=DF, DF=EF.
Therefore, PA=√2*EF.
Therefore, PC = √ 2 * CF = √ 2 * (CE+EF) = √ 2 * CE+√ 2 * EF = √ 2 * CE+PA.
That is, the relationship among PC, PA and CE is: PC=√2CE+PA.
2. If point P does not coincide with O and C on the OC line, PE is perpendicular to PB, and the intersection CD of PE is at point E. Please complete Figure 3 and judge whether the question 1.2 is valid. If not, write the corresponding conclusion.
As shown on the right
With the above ideas.
Because BC⊥CE's PB⊥PE
So, B, P, C and E are four * * * cycles.
Therefore ∠PEC=∠PBC.
However, in △PBC and △PDC:
BC=DC (known)
∠ PCB =∠ PCD = 45 (certification)
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So, △ PBC △ PDC (SAS)
Therefore, ∠PBC=∠PDC.
Therefore ∠PEC=∠PDC.
And PF⊥DE
So, DF=EF
Same as above:
PA=√2*PG=√2*DF=√2*EF
PC=√2*CF
Therefore, pa = √ 2 * ef = √ 2 * (ce+cf) = √ 2 * ce+√ 2 * cf = √ 2 * ce+PC.
That is, the relationship among PC, PA and CE is: PA=√2*CE+PC.