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.
Continue reading at:
gogeometry.com/ArchBooLem10.htm
Tuesday, December 16, 2008
Archimedes' Book of Lemmas, Proposition #10
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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.
ReplyDeleteThis 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.
< BCA = < CEB = < DFE = < DFA
ReplyDeleteHence ABFC is cyclic and so the above angles also = < CFA = < FCE
So Tr.s EFG & CFG are congruent and EG = GC
Sumith Peiris
Moratuwa
Sri Lanka