Verification: (1)PQ2=PO2+QO2-2R2
(2)
O is the center of △PQM.
Proof: (1) Prove ∠ PDQ > ∠ PQC first.
this is because
∠PDQ =∠DQC+∠DCQ >∠DQC+∠ABC =∠DQC+∠ADQ >∠DQC+∠DQP
=∠DQC+∠ADQ=∠PQC
In this way, you can do ∠PDQ ∠PDE=∠PQC, and E is on PQ.
therefore
d、E、Q、C
Four o'clock * * * circle.
And ∠DEQ=∠BCD=∠DAP.
therefore
D, E, P and A are also four-point * * * cycles.
According to the circular power theorem
QP QE=QD QA=(QO+R)(QO-R)=QO2-R2
In the same way.
PQ PE=PD PC=PO2-R2
Add two types, PQ.
(PE+EQ)=PO2+QO2-2R2
that is
PQ2=
PO2+QO2-2R2
(2) Let the extension line of the circumscribed circle of A, D A, D, M intersect QM at f,
Then a, d, m and f are four-point circles.
∠QBD=∠QAM=∠DFM，
therefore
B, f, d and q are also four-point circles.
According to the circular power theorem
QM QF=QD QA=QO2-R2
MQ MF=MB MD=R2-MO2
The two formulas are subtracted, QM2=QO2+MO2-2R2.
In the same way.
PM2=PO2+MO2-2R2
Subtract these two formulas, QM2-PM2=QO2-PO2.
therefore
OM⊥PQ
In the same way; In a similar way
OP⊥QM,OQ⊥PM
therefore
O is the center of △PQM.
2.
(1997 China Mathematical Olympics) The quadrilateral ABCD is inscribed in a circle, and its sides AB and DC extend to point P, and AD and BC extend to point Q, where Q is two tangents QE and QF of the circle, and the tangents are E and F respectively. It is proved that the straight line of p, e and f is * * *.
Prove:
Connect PQ, make ⊙QDC, and cross PQ at point M,
then what QMC=？ CDA=？ CBP, so M, C, B and P are four * * * cycles.
pass by
PO2-r2=PC PD=PM PQ，
QO2-r2=QC QB=QM QP，
The two expressions are subtracted to get PO2-Qo2 = PQ (PM-QM).
=(PM+QM)(
PM-QM)=PM2-QM2，
∴
OM⊥PQ.
∴
O, f, m, q and e are five-point circles.
Even PE, if PE crosses ⊙O at F 1 and ⊙OFM at F2, then
For ⊙O, there are pf 1 PE = PC PD,
For ⊙OFM, there is PF2 PE = PM PQ = PC PD.
∴
Pf 1 PE = PF2 PE, that is, f 1 and F2 coincide at the common point f of two circles, that is, the three-point line of p, f and e. 。
3. As shown in the figure, the center of the semicircle is O and the diameter is AB. A straight line intersects the semicircle at C, D and AB and at M..
(MB