Archimedes (287–212 BC, ancient Greece) was one of the greatest mathematicians of all time. Together with Newton and Gauss, he is known as the three mathematicians.
< h1 class="pgc-h-arrow-right" data-track="2" > theorem definition</h1>
As shown in the figure on the right, AB and BC are the two strings of ⊙O (i.e., ABC is a folded string of the circle), BC > AB, M is the midpoint of the arc ABC, and the vertical foot D of the vertical line made from M to BC is the midpoint of the folded string ABC, that is, CD = AB + BD.
Definition: A polyline of two strings starting from any point on the circumference of the circle, which we call a folded string of the figure.
<h1 class="pgc-h-arrow-right" data-track="5" > validation derivation</h1>
Method 1: Complement the short method 1
As shown, extend DB to F so that BF=BA
∵ M is the midpoint of arc ABC
∴∠MCA=∠MAC=∠MBC
∵ MBAC four-point rounding
∴∠MCA+∠MBA=180°
∵∠MBC+∠MBF=180°
∴∠MBA=∠MBF
∵MB=MB,BF=BA
∴△MBF≌△MBA
∴∠F=∠MAB=∠MCB
∴MF=MC
∵MD⊥CF
∴CD=DF=DB+BF=AB+BD
Method 2: Complement the short method 2
Extend AB to E so that BE=BD
∵M is the midpoint of arc AB,
∴∠MBC=∠MAC=∠MCA
∵ M, B, A, C are four-point syntactic
∵∠MBE+∠MBA=180°
∴∠MCA=∠MBE
∴∠MBC=∠MBE
∵BE=BD,MB=MB
∴△EBM≅△DBM
∴∠E=∠MDC=90°,ME=MD
又∵MA=MC
∴△MEA≅△MDC
∴DC=AE=AB+BE=AB+BD
Method 3: Truncation method 1
As shown in the figure, DG=DB is intercepted on a CD
∵MD⊥BG
∴MB=MG，∠MGB=∠MBC=∠MAC
∴∠MAC=∠MCA=∠MGB
即∠MGB=∠MCB+∠BCA=∠MCB+∠BMA
又∠MGB=∠MCB+∠GMC
∴∠BMA=∠GMC
∵MA=MC
∴△MBA≌△MGC(SAS)
∴AB=GC
∴CD=CG+GD=AB+BD
Method 4: Truncation method 2
As shown in the figure, CG=AB is intercepted on a CD
∴MA=MC
∵∠BAM=∠BCM
∴MB=MG
∴BD=DG
<h1 class="pgc-h-arrow-right" data-track="62" > method 5: perpendicular method</h1>
As shown in the figure, MH⊥ ray AB, the vertical foot is H.
∵MD⊥BC
∴∠MDC=90°=∠H
∵∠MAB=∠MCB
∴△MHA≌△MDC(AAS)
∴AH=CD，MH=MD
又∵MB=MB
∴Rt△MHB≌Rt△MDB(HL)
∴HB=BD
∴CD=AH=AB+BH=AB+BD
Method 6: Rounded angle method
Extend the MD intersection O to E, connect EC, EA,
Extend EA cross CB extension line at F.
∵ M is the AMC midpoint
∴∠1=∠2
∵ MD⊥BC
∴∠EDF=∠EDC=90°,
∵ED=ED
∴△EDF≅△EDC
∴∠C=∠F,DF=DC
∵ A.B, C, E four-point syntactic
∴∠C+∠BAE=180°.
And ∠3 + ∠BAE = 180°
∴∠C=∠3
∴∠F=∠3
∴BF=AB
∴CD=FD=BF+BD=AB+BD
Example 1 ☆☆☆☆☆
As shown in the figure, the known points A, B, C, D are in order on the circle O, AB = BD, BM⊥AC, and the vertical is M.
Proof: AM=DC+CM.
1. (★★☆☆☆) As shown in the figure, the known points A, B, C, D are in order on the circle O, AB =BD, BM⊥AC, the vertical foot is M. If AM=5, CM=1, then CD=_________
2. As shown in the figure, it is known that in △ABC, D is the point above the AC, and AD = DC + CB, and the dot D is the perpendicular junction circle of the AC at the point M.
3. As shown in the figure, △ ABC is connected to ⊙O, BC = 2, AB = AC, point D is the moving point on the arc AC,
and cos∠ ABC=
(1) Find the length of AB.(2) Find AD· The value of AE.
(3) Pass the A point as AH⊥ BD, verify: BH= CD + DH.