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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 639.
Connect AG Per the result of problem 636, we have AG perpen. To DH.Triangle CDH congruence to triangle GDA ( Case ASA)So AD= DHPeter Tran
Let AC = 2a, BC = 2b, FG = c and FH = xThen 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 = DHSumith PeirisMoratuwaSri Lanka
Some clarifications on my above proofIf M is the foot of the perpendicular from F to AB then we can show that OM = b+c-a. This implies that OD = aAbove I have written the Pythagorean expressions twice for FM