(1) Since it is a multiple-choice question, you can use many methods. In my opinion, if the simplest method of multiple-choice questions is the limit method, the right angle ∠BCD rotates around point C by an angle θ (0

CQ？ +CP？ =CD？ +CB？ According to the rectangle: CQ? +CP？ =CD？ +CB？ =AD？ +AB？ Because of the three vertical theorems, DF=AD.

AB=BE。 So: CQ? +CP？ =CD？ +CB？ =AD？ +AB？ =DF？ +PE？ Because point Q is point D and point P is point B, CQ? +CP？ =CD？ +CB？ =AD？ +AB？ =DF？ +PE？ =QF？ +PE？ Which is CQ? +CP？ =QF？ +PE？ ,

(2) Geometric method:

Extend PC to point m so that PC=CM. Then connect MQ and FM (in fact, do the symmetry of △PCE about point C △MFC).

Reconnect PQ

Because △PCE is symmetrical about point C △MFC, △ PCE △ MFC. So: PE=FM. PC=MC。 ∠PCE=FCM。

∠PEF=∠MFQ

Because ∠BCD rotates around point C by an angle θ (0

Because ∠ BCD = ∠ QCB+∠ QCD = 90, then ∠ QCP = ∠ QCB+∠ BCP = 90, then QC is the height of △MPQ, and PC=MC, then: △ QMC △

Because ∠ PEF = ∠ MFQ, ∠ QFE+∠ PEF = ∠ QFE+∠ MFE = 90, PE=FM.

So: QM? =QF？ +FM？ =QF？ +PE？

QP？ =CQ？ +CP？

Because: QM=QP

So: CQ? +CP？ =QF？ +PE？

(3) Let a rectangle have a length and a width, and establish a rectangular coordinate system with C as the origin, CB as the X axis and CD as the Y axis.

Then: q (btanθ, b) f (-a, b) p (a, -atanθ) e (a, -b)

CQ？ +CP？ =(btanθ)？ +b？ +a？ +(-atanθ)？ =(a？ +b？ )(tanθ？ + 1)

QF？ +PE？ =(-a-btanθ)？ +(-b-atanθ)？ =(a？ +b？ )(tanθ？ + 1)

So: CQ? +CP？ =QF？ +PE？