Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #9

Problem 648: Perpendicular chords and arcs
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #9 (high school level) and lift up your geometry skills.


Archimedes' Book of Lemmas #9.
Continue reading at:
gogeometry.com/ArchBooLem09.htm

Posted by Antonio Gutierrez at 3:36 PM
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5 comments:

  1. AnonymousJune 1, 2009 at 9:02 AM

    Here we go:
    The statement to prove is equivalent to the statement that arc AD+arc CB is the length of half the circumference of the circle. We will prove the last one.

    Draw the line l, which passes through the center of the circle and is parallel to AB. Now consider the reflection of the setting in l. The line of the circle is mapped to itself and now the image of A is named A', the image of B is B' etc. Now, as l is the perpendicular bisector of CD, we have D'=C and C'=D.

    As AA' is perpendicular to the mirror l, it must also be perpendicular to its parallel AB. But when A'AB is a right angle on the circle line, its respective chord A'B has to be a diameter by the converse of Thales' theorem.

    So, we have, that
    0.5 diameter=arc A'B=arc A'D'+arc CB= arc AD+arc CB, as claimed. (by Thomas)

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  2. Peter TranJuly 30, 2011 at 7:42 AM

    http://img841.imageshack.us/img841/5839/problem648.png
    Connect AC and BD
    Note that angle(APC)=angle(PDB)+angle(PBD)= 90
    and angle(DPB)=angle(ACP)+angle(CAP)=90
    So Arc(AC)Arc(BD)=Arc(AD)+Arc(BC)
    Peter Tran

    ReplyDelete
  3. PravinJuly 30, 2011 at 8:20 AM

    Let O be the center of the circle
    ∠APC = 90°
    ∴ ∠ACP + ∠CAP = 90°
    Same as ∠ACD + ∠CAB = 90°
    ∴ ∠AOD + ∠COB = 180°
    ∴ measure of arc AD
    + measure of arc BC = 180°
    Similarly the other sum = 180°

    ReplyDelete
  4. PrabhakarNovember 16, 2011 at 6:54 AM

    Construction=join AC
    1.ang. CAB=1/2 arc BC.......(inscribed ang.thm(I.A.t))
    2.ang. ACD=1/2 arc AD.......(inscribed ang.thm(I.A.t))
    3.hence,
    ang. CAB + ang. ACD =m 1/2(arc BC+ arc AD)....(1+2)
    4.in tri ACP,
    ang. CAB + ang. ACD =ang CPB....(ext ang of triangle)
    5.hence,
    ang CPB = m 1/2(arc BC+ arc AD)....(from 4,5)
    6.90=m 1/2(arc BC+ arc AD)
    7.180=m (arc BC+ arc AD)
    8.similarly by constructing BC we can get,
    ang. APD=m 1/2(arc AC+ arc BD)
    9.hence
    180=m(arc AC+ arc BD)
    10.hence,
    m (arc BC+ arc AD)=m(arc AC+ arc BD)......(from 7 , 10)
    hence proved.

    ReplyDelete
  5. c.t.e.o.April 10, 2023 at 1:14 PM

    https://photos.app.goo.gl/jJqSqY7kBqDctKzt9

    ReplyDelete

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