Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #2

Problem 641: Diameter, tangents, perpendicular
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #2 (high school level) and lift up your geometry skills.

Archimedes' Book of Lemmas #2.


  1. Let M be the center of the circle.
    Extend BC and AD and let them intersect in L.
    Let MC and BD intersect in K.

    Then MBCD is a kite (two tangent rays, two radii), so its diagonal MC is the perpendicular bisector of its diagonal BD. So MKB=90°. We also have ADB=90° (Thales) and as orthogonals to the same line, MC and DL are parallel. Therefore BLD and BCK are similar, and we have BC/CL=BK/KD=1, or BC=CL.

    Now ED and BL are parallel (orthogonal to AB), and by the intercept theorem CL/FD=AC/AF=BC/EF, and if we drop the middle part, we get EF/FD=BC/CL=1, which was claimed. (by Thomas)

  2. bjhvash44@sbcglobal.net

    use anom's construction to id point L
    ang DAB=CDB=CBD= 1/2 arc DB=a
    angALB=complement ofDAB =b
    angLDC=complement of a=b
    trian CLD is isosceles
    AC is median of trian ALB It bisects all parallels to BL

  3. http://img109.imageshack.us/img109/5805/problem641.png
    Extend AD to G ( G on BC, see picture)
    Triangle BDG is a right triangle ( angle ADB =90)
    Triangle DCB is isosceles ( DC=CB)
    Triangle CDG is isosceles ( angle CDG= angle DGC= 90- angle CBD)
    So CD=CB=CG
    Since DE// BG and C is midpoint of BG so F is the midpoint of DE
    Peter Tran

  4. If AD, BC meet at G and DE, OC meet at H then it can also be shown that BCHD is a parellogram and that BCDH is a rhombus.

  5. Triangles ADE and ABD are similar, AC is symmedian of triangle ABD and consequently it is median of triangle ADE, done.