Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 639.
Saturday, July 23, 2011
Problem 639: Semicircle, Diameter, Perpendicular, Inscribed Circle, Tangent, Arbelos, Congruence
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Connect AG
ReplyDeletePer the result of problem 636, we have AG perpen. To DH.
Triangle CDH congruence to triangle GDA ( Case ASA)
So AD= DH
Peter Tran
Let AC = 2a, BC = 2b, FG = c and FH = x
ReplyDeleteThen 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
Some clarifications on my above proof
ReplyDeleteIf 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