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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below for more details.
Let AO meet the circumcircle at N1)Then < HCB = 30 = < CBN since AN is a diameter. So HC//BNAlso < BAN = 90-C = < BCN = < CBH so BH //NCSo BHCN is a parellelogramNow Tr. ONC is equilateral so NC = BH = BO = R where R is the circumradius 2) Easily < CAN = 30 = <ACO hence < AOC = 120 = AHC. So AOHC is cyclic and so < AHD = 30. So < BDE = 60 and hence Tr. BDE is equilateral 3) This implies that Tr.s BDH and BOE are congruent ASA hence DH = OE and so DO = HE 4) Let BD = DE = BE = p and if AH meets BC at M let ME = q and let CH meet AB at SBS = a/2 since Tr. BCS is 30-60-90. So DP = p - a/2 = DH/2 hence DH = 2p -a and so HE = p - (2p - a) = a - p = DO Therefore OH = 3p - 2aNow q = HE / 2 = (a-p)/2 and BM = c/2 = p - q. So c/2 = p - (a-p)/2 and hence 3p = c + a. (1)So AD + CE = c-p + a-p = p from (1 ) above and is therefore = to BD5) AB -BC = c-a = (c-p) - (a-p) = AD - CE = a + c -2a = 3p - 2a = OH Sumith PeirisMoratuwaSri Lanka
See problem 1062.
Ajit sir I need help for the following problemhttp://www.artofproblemsolving.com/community/u271471h1157160p5494288plz reply me at firstname.lastname@example.org
I've solved the problem using physics.
GG Problem 1161. Let X be the midpoint of AC.BH = 2OX = 2R cos B = R = BO.So <BOH = <BHO ......(i).Extend BH to cut AC at K and circle at L.Extend BO to cut the circle at M.Connect CM.<ABM = 90 - <AMB = 90 - <ACB = 90 – C = <KBC.(i.e)<DBO = < EBH ....(ii).Subtract (ii) from (i) to get<BDE = <BED = 60 and ΔBDE is equilateral.Note that ΔBDO and ΔBEH are congruent andADH is a 30-120-30 triangle.So AD = DH.Similarly EC = EH.Hence AD + CE = DH + EH = DE = DB