Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #13

Problem 653: Diameter, chord, perpendicular
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #13 (high school level) and lift up your geometry skills.

Click the figure below to see the complete problem 653.

Archimedes' Book of Lemmas #13.
Continue reading at:
gogeometry.com/ArchBooLem13.htm

Posted by Antonio Gutierrez at 3:42 PM
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7 comments:

  1. AnonymousJune 11, 2009 at 9:26 AM

    The situation becomes clear, if you extend the lines AF and BG, so that they intersect the circle a second time.

    Let then AF intersect the circle in P and let BG intersect the circle in Q. Then we have angles BPA=AQB=90° (Thales). As AP and BQ are parallel, QBA=PAB and ABP=BAQ as they are alternate angles; and as the sum of them is 180° (they are part of a cyclic quadrangle), each of them is 90°.

    Thus APBQ is a rectangle. If we now draw the line through the center of the circle, which is perpendicular to CD, this will be a median of the rectangle (as a parallel to two sides which passes through the center). So, if we consider a reflection in this line, the circle and the rectangle are invariant and the segments DF and CG are mapped to each other. Thus they are equal. (by Thomas)

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  2. Peter TranJuly 31, 2011 at 11:34 PM

    Let O is the center of circle and I is the projection of O over CD.
    Since O is the midpoint of AB so I is the midpoint of FG and DC( F, I, G are the projection of A,O,B)
    CG=CI-IG
    DF=ID-IF
    Since ID=IC and IF=IG so CG=DF
    Peter Tran

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  3. AjitAugust 4, 2011 at 10:16 PM

    AD = ABcos(BAD) & DF=ADcos(ADC) or DF=ABcos(ADC)cos(BAD) But /_ADC=/_ABC (angles in the same sector) & /_BCD=/_BAD for the same reason. Hence DF=ABcos(ABC)cos(BAD)=BCcos(BCD)=CG
    Ajit

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  4. PravinAugust 7, 2011 at 4:27 AM

    Draw OH ⊥ CD. OH bisects CD
    CH = HD
    AF ∥ OH ∥ BG
    Transversal AB is bisected at O
    So transversal GF is bisected at H
    GH = HF
    Follows
    CG = CH - GH = HD - HF = FD

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  5. PradyumnaSeptember 8, 2011 at 7:36 AM

    produce BG to cut the circle at P,=>angle BAP=90
    If we observe AFPG, angles P,G,F are 90 degrees
    =>angle FAP is 90 degrees,therefore AFPG is rectangle =>AF=PG
    Now triangles AFD and PGC are similair and AF=PG
    =>triangles AFD and PGC are congruent
    =>CG=FD

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  6. AnonymousJune 25, 2013 at 4:13 PM

    http://img43.imageshack.us/img43/8577/u2mk.png

    ReplyDelete
  7. Sumith PeirisOctober 23, 2016 at 12:07 AM

    Let AF meet the circle at H.

    DHBC is an isoceles trapezoid and HFGB is a rectangle.

    Triangles BGC & HFD are therefore congruent ASA.

    Hence DF = CG.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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