Saturday, July 23, 2011

Problem 639: Semicircle, Diameter, Perpendicular, Inscribed Circle, Tangent, Arbelos, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 639.

Online Geometry Problem 639: Semicircle, Diameter, Perpendicular, Inscribed Circle, Tangent, Arbelos, Congruence.

3 comments:

  1. Connect AG
    Per the result of problem 636, we have AG perpen. To DH.
    Triangle CDH congruence to triangle GDA ( Case ASA)
    So AD= DH
    Peter Tran

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  2. Let AC = 2a, BC = 2b, FG = c and FH = x

    Then from similar Tr.s we can show that x = c(b+c)/(b-c)

    Applying Pythagoras twice to Tr. CDF,

    (a+b-c)^2 - (a+c-b)^2 = (b+c)^2 - (b-c)^2 which simplifies to a = bc/(b-c)

    So DH = x+b+c = b(b+c)/(b-c)

    DA = 2a+b which is again = to b(b+c)/(b-c) by substituting a = bc/(b-c)

    So DA = DH

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  3. Some clarifications on my above proof

    If M is the foot of the perpendicular from F to AB then we can show that OM = b+c-a. This implies that OD = a

    Above I have written the Pythagorean expressions twice for FM

    ReplyDelete