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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
through D draw line perpendicular to AB, end on F;Connect GC,triangle GDF similar to triangle DGE, ( angle AGD = angle GCD)a is radius of circleso X^2 = 2a(a+b)
http://s29.postimg.org/m7r8fv3hz/pro_1083.pngDraw lines and points per sketchLot O is the center of the circle, we have FO⊥AE and GO⊥AB => AFOG is a squareSo CD= 2.AF= 2aWe also have ∠ (HDG) =∠ (BGC) and ∠ (CDG)= ∠ (BGC)so GD is an angle bisector of angle CDHDH=GK= a + bRelation in the right triangle DGC give DG^2= DK.DCor x^2=2a.(a+b)
May I point out a typo ? Instead of DH=GK= a + b, should it not be DH=DK= a + b ?
Thanks for the correction. the corrected statement is " DH=DK=a+b"Peter Tran
https://www.dropbox.com/s/uggeh3qzew0fdld/1083.fig?dl=0Complete the rectangle BAEJ and semicircle to full circle.Let EJ and diameter GOK intersect at H.Then x^2 = GD^2 = GK.GH = 2a(a + b).
Let H be on AG such that HD//BC//GO where O is the centre of the semicircle CGDAGOF is a square of side a and the radius of the semicircle is aTr. a CDG and GDH are similar hence2a/x = x/(a+b) from which the result followsSumith PeirisMoratuwaSri Lanka
Further if DE = c and DF = y then it can be shown that y^2 = 2ac
Moreover if J is the food of the perpendicular from G to CD, CJ = a-b and GJ^2 = a^2 - b^2