Current location - Training Enrollment Network - Mathematics courses - Mathematical pm2
Mathematical pm2
1. As shown in the figure, the opposite sides BC and AD inscribed in the extended quadrilateral ABCD, BA and CD intersect at P and Q respectively, and the diagonal AC and BD intersect at M, where O is the center of the circumscribed circle of the quadrilateral ABCD and R is the radius.

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