Four important theorems:
Menelaus Theorem (Mei's Line)
△ If there are points P, Q and R on the three sides BC, CA and AB of△ △ABC or its extension line, the necessary and sufficient conditions for straight lines P, Q and R*** are as follows.
Ceva Theorem (Ceva Point)
△ABC has points P, Q and R on three sides of BC, CA and AB, then the necessary and sufficient conditions for points AP, BQ and CR*** are as follows.
ptolemy's theorem
The necessary and sufficient condition for the sum of the products of two opposite sides of a quadrilateral to be equal to its diagonal product is that the quadrilateral is inscribed with a circle.
Siemsen Theorem (siemsen Line)
The necessary and sufficient condition for drawing a perpendicular from a point to three sides of a triangle is that the point falls on the circumscribed circle of the triangle.
Example:
1. Let AD be the center line of the side BC of △ABC, and the straight line CF intersects with AD at F, which proves that:
Analysis of CEF section △ABD→ (Meyer theorem)
The commentary can also add auxiliary lines to prove that one of A, B and D is a parallel line of CF.
2. The straight lines passing through the gravity center G of △ABC intersect with CB in AB, AC, F and D in E respectively.
Prove:
Analyze the link and extend the intersection of AG and BC to m, then m is the midpoint of BC.
DEG section △ABM→ (Meyer theorem)
DGF section △ACM→ (Meyer theorem)
∴=== 1
Comment on Meyer theorem
3.d, E and F are respectively beside BC, CA and AB of △ABC.
, AD, BE and CF cross into △LMN.
Find LMN.
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Comment on Meyer theorem
4. Based on the sides of △ABC, make similar isosceles △BCE, △CAF and △ABG outward. Prove that AE, BF and CG intersect at one point.
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Comments on Seva Theorem
5. It is known that ∠B=2∠C in △ABC. Verification: AC2 = AB2+ AB BC.
It is analyzed that parallel lines with A as BC intersect △ABC, and the circumscribed circle is D, which connects BD. Then CD=DA=AB, AC=BD.
According to Ptolemy theorem, AC BD = AD BC+CD AB.
Comment on Ptolemy Theorem
6. The regular heptagon A1A2A3A4A5A7 is known.
Prove: (The 2nd1All-Soviet Mathematics Competition)
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Comment on Ptolemy Theorem
7. The extension line of the BC side height AD of 7.△ ABC circumscribes P, PE⊥AB is in E, and the extension line of ED is in F. ..
Verification: BC ef = BF ce+be cf.
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Comments on siemsen's Theorem (siemsen Line)
8. Diagonal AC and CE of regular hexagon ABCDEF are divided into AM: AC = CN: Ce = K by straight lines of inner points M and N, and B, M and N*** respectively. Find K. (23-IMO-5)
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Comment area method
9.O is a point within △ABC, the distances from O to BC, CA and AB are represented by da, db and dc respectively, and the distances from O to A, B and C are represented by Ra, Rb and Rc.
Verification: (1) a ra ≥ b db+c dc;
(2)a Ra≥c d b+ b DC；
(3) Ra+Rb+Rc≥2(da+db+dc).
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Comment area method
At 10. △ ABC, H, G and O are vertical center, center of gravity and outer center respectively.
Verification: H, G, O three-point * * * line, HG=2GO. (Euler line)
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Comment on the same law
In 1 1. △ ABC, AB=AC, AD⊥BC = D, BM, BN ∠ABC, intersecting with AD in M and N, and extending CM and AB in E.
Verification: MB//NE.
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Comment symmetry transformation
12.g is the center of gravity of △ABC. With AG as the chord, BG is tangent to G, and the intersection of CG extends to D ... Verification: AG2 = GC GD.
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Comment translation transformation
13.c is the point above ⊙O with diameter AB=2, and p is within △ABC. If the minimum value of PA+PB+PC is 0, find the area s of △ABC at this time.
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Annotation rotation transformation
Fermat point: O is a point within △ABC, ∠ AOB = ∠ BOC = ∠ COA =120; P is any point in △ABC. Proof: PA+PB+PC≥OA+OB+OC. (O is fermat point)
Analysis will cc', OO', PP', link OO', PP'. Then △B OO' and △B PP' are regular triangles.
∴OO'=OB,PP' =PB. Obviously △ bo' c' △ BOC, △ BP' c' △ BPC.
Because ∠ bo' c' = ∠ BOC =120 =180-∠ bo' o, ∴A, o, o', c' four-point * * line.
∴ AP+PP'+p' c' ≥ AC' = ao+oo'+o' c', that is, PA+PB+PC≥OA+OB+OC.
14. (National Competition' 95) The inscribed circle O of rhombic ABCD intersects with all sides at E, F, G and H respectively, and the tangent of ⊙O on arc EF and arc GH intersects with AB, BC, CD and DA at M, N, P and Q respectively.
Authentication: MQ//NP.
Analysis from AB‖CD: To prove MQ‖NP, you only need to prove ∠AMQ=∠CPN.
Combining ∠A=∠C, you only need to prove it.
△AMQ∽△CPN
←，AM CN=AQ CP .
Connect AC and BD, the intersection point is the center o OF inscribed circle, let MN and ⊙O cut at k, and connect OE, OM, OK, on and of. Remember ∠ABO=φ, ∠MOK=α, ∠KON=β, then
∠EOM=α，∠FON=β，∠EOF=2α+2β= 180 -2φ.
∴∠BON=90 -∠NOF-∠COF=90 -β-φ=α
∴∠cno=∠nbo+∠nob=φ+α=∠aoe+∠moe=∠aom
And ∠ocn =∠ Mao, so,
∴AM CN=AO company
In the same way, AQ CP = ao company
To annotate ...
15. (National Competition' 96) ⊙O 1 and ⊙O2 are all tangent to the straight lines on the three sides of Δ δABC, with E, F, G and H as tangents, and the extension lines of EG and FH intersect with P. Proof: PA⊥BC.
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To annotate ...
16.(99 National Games) As shown in the figure, in the quadrilateral ABCD, the diagonal AC bisector ∠BAD. Take a little e on the CD, BE and AC intersect at f, and extend DF to BC and G. Verify: ≈GAC =≈EAC.
Prove: by linking BD with AC and H, and using Seva theorem for △BCD, we can get
Because AH is the angle bisector of ∠BAD, from the angle bisector theorem,
Therefore, it is available.
The parallel line of AB passing through C intersects the extension line of AG at I, and the parallel line of AD passing through C intersects the extension line of AE at J.
Then,
Therefore, CI=CJ.
And because CI//AB, CJ//AD, ∠ACI=π-∠BAC=π-∠DAC=∠ACJ.
Therefore, △ ACI △ ACJ, thus ∠IAC=∠JAC, that is ∠GAC=∠EAC.
It is known that AB=AD, BC=DC, AC and BD intersect at O, and any two straight lines EF and GH passing through O intersect with four sides of quadrilateral ABCD at E, F, G and H. Add GF and EH, and pay BD to m and n respectively. Verification: OM = on. (5th CMO)
Proof: If △ E'H' △ E' OH', you only need to prove e', m, H' * * *, that is, the points of E'H, BO and GF.
Remember ∠BOG=α, ∠GOE'=β. To connect e' f to BO to K, just prove = 1(Ceva inverse theorem).
=== 1
Note: Zheng shape: a quadrilateral with one diagonal vertically bisecting the other diagonal.
Corresponding to the three lines of 2: ∠ e 'ob = ∠ FOB, e 'h, GF and BO in the 99 League, they are * * *. Proof: ∠ gob = ∠ h 'ob.
In fact, the above conditions are all necessary and sufficient conditions, and the conclusion is still valid when m is on the OB extension line.
Proof method: the same method.
Butterfly theorem: P is the midpoint of the chord AB of ⊙O, and two chords CD and EF of ⊙O are introduced through point P to connect DE and AB in M and CF and AB in N. Prove: MP=NP.
Let GH be the diameter of p, FF'F, obviously' ∑⊙o. p∈GH, ∴ PF' = PF. ∵ PFPF', PAPB, ∴∠ FPN = ∠ F 'PM, PF = PF'.
And FF'⊥GH, AN⊥GH, ∴ff' ab. ∴∠F'PM+∠MDF'=∠FPN+∠EDF'
= ∠ eff'+∠ EDF' = 180, ∴P, m, d, f' four-point * * * circle. ∴∠PF'M=∠PDE=∠PFN。
∴△PFN≌△PF'M,PN=PM。
The general conclusion of the commentary is: given a point P on the chord AB in a ⊙O with radius R, two intersecting chords CD and EF pass through P, even CF and ED intersect AB at M and N, given OP=r and the distance from P to the midpoint of AB is A, then. (Analytic proof: using the knowledge of conical system)