Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #10

Problem 649: Tangents, secant, chords, parallel, perpendicular
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #10 (high school level) and lift up your geometry skills.

Archimedes' Book of Lemmas #10.
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  1. Let secant AD meet the circle in I and let O be the centre of the circle. Join B & C to O and F to C. Since AB & AC are tangents to the circle, points O, C, A & B are concyclic –--(1). Now /_ABC=/_BEC (Alternate segment Theorem) and /_BEC=/_BFA (AD//CE) or /_BFA=/_ABC.
    This means that A, B, F & C are concyclic once again by the aforementioned theorem---(2). In (1) & (2) three points viz. A, B & C are common and thus A,B,F,O and C are all on the same circle of which AO is the diameter since /_ABO=90 deg.
    Therefore, /_AFO also =90 deg. Or F is the midpoint of the chord DI.
    Since DI//EC and FG perpendicular to CE, G is the midpoint of CE. QED.

  2. < BCA = < CEB = < DFE = < DFA
    Hence ABFC is cyclic and so the above angles also = < CFA = < FCE
    So Tr.s EFG & CFG are congruent and EG = GC

    Sumith Peiris
    Sri Lanka