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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1174.
Let the bisector of < ABC meet AC at E, CD at F and AD at G.For ease of typing I use @ for alpha and $ for theta Easily < ABO = OBC = @/2 and < CBE = @ so that < EBD = 2@Also since BG is the perpendicular bisector of AD, < BAE = EAF = BDE = EDF = $So C is the incentre of Tr. ABF so < DFG = < AFG = < AFC = < CFE and each of these must therefore be essentially 60 degrees since the last 3 of these 4 angles add upto 180.Hence from Tr. BDF, 2(&+$) = 60 and so &+$ = 30.....(1)In Tr. ABC, OA is an external angle bisector at A so x+@/2 = 90 -$/2 from whichx + (@+$)/2 = 90 Substituting @+& = 30 from (1) thereforex+30/2 = 90 and hence x = 75Sumith PeirisMoratuwaSri Lanka
http://s17.postimg.org/59873mej3/pro_1174.pngLet angle bisector of angle ABD meet CD at F and AD at EObserve that due to symmetry we have∠(BAF)= ∠ (BDF)=2. ThetaAnd tri. AFD is isoscelesAC and BC are angles bisectors of ∠ (BAC) and ∠ (CFA) So ∠ (BFC)= ∠ (CFA)=2.alpha+2.theta => ∠ (FDA)=alpha+thetaIn right tri. BED we have alpha+theta=30We have∠ (HOA)= theta/2 and ∠ (ABO)=alpha/2In right tri. HBO we have x+theta/2+alpha/2= 90 => x= 75