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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 788.
http://img687.imageshack.us/img687/6215/problem788.pngDraw lines per attached sketchWe have <BDA=<BFA=.5*<BQA=<BQOAnd <BEA=<BCA=.5*<BOA=<BOQSo all 3 triangles OBQ, EBF and CBD are similar per case AATri. CBD is the image of Tri. EBF in the spiral similarity transformation centered B , angle of rotation = theta=<EBC=<FBD=<FADSince G is the image of H in this transformation so <HBG=<FAD= thetaSo quadrilateral AHBG is cyclic
Join BC, BD, BE and BF. Angle AFB = angle ADBAngle AEB = angle ACBThus, triangle BCD ~ triangle BEF. Now rotate triangle BCD to triangle BEF, with center B, such that BC and BE coinside, BD and BF coinside, with different lengths. Then CD becomes EF (with different lengths), BG becomes BH (with different lengths). Therefore, the rotation angle isangle GBH, and also same as angle CAE. Hence, AHBG is concyclic.
Let M be the midpoint of OQ. Join OA, QA, MBPower of H w.r.t. circle (O) = HA.HE = OA² - OH²Power of H w.r.t. circle (Q)= HA.HF = HQ² - QA²So OA² - OH² = HQ² - QA² (since HE = HF),OA² + QA² = OH² + HQ² which is same as 2AM² + MQ²/2 = 2HM² + MQ²/2Follows MA = MHSimilarly equating powers of G w.r.t. circle (Q),it can be shown that MA = MGOQ being the perpendicular bisector of AB,we have MA = MBHence MA = MH = MG = MB and A, H, B, G are concyclic.
Let CG = GD = p and let EH = HF = qTr.s BCD and BFE are similar so, 2p/2q = s/t where BC = s and BE = t Hence s/p = t/qSo in Tr. s BEH and BCG < E = < C and the sides which hold these angles are proportional Hence
...the triangles are similar and the result follows Sumith PeirisMoratuwaSri Lanka
Problem 788Is <ACB=<AEB,<ADB=<AFB then triangleCBD similar triangleEBF.So BD/BF=CD/EF=2GD/2HF=GD/HF.Then triangleBDG similar triangleBFH .So <BGD=<BHF=<BHA.Therefore AHBG is cyclic.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE