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Proposed ProblemSee also: Complete Problem 30, Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
From F draw FD perpendicular to AB meeting the latter in D. Now ED = r, angle EAB = angle EAF = A/2, r = AE sin(A/2) and AE = AFcos(A/2) which gives us: r/sin(A/2) = AFcos(A/2) or AF = r/sin(A/2)cos(A/2) = 2r/sin(A). Finally, BG = FD = AFsin(A) = 2r/sin(A) * sin(A) = 2r.Ajit: firstname.lastname@example.org
Let G be an intersecting point of the segment AB and the extension of FE and letP,Q and R be ponts on segments AB, BC and FG, respectively with EPㅗAB, EQㅗBC and ERㅗFG.Then we see that BQ=EP=EQ=r and two triangles EPG and ERF are congruence.Hence we get BG=BQ+QG=BQ+ER=BQ+EP=2r.
Let G be an intersecting point of the segment AB and the extension of FE and letP,Q and R be ponts on segments AB, BC and FG, respectively with EPㅗAB, EQㅗBC and ERㅗFG.Then we see that BQ=EP=EQ=r andtwo triangles EPG and ERF are congruence.Hence we get BG=BQ+QG=BQ+ER=BQ+EP=2r
Through E draw MN ∥ AB with M on AF and N on BG.∠AEM = ∠EAB = A/2 -> ME = MA∠MEF = 90° - A/2 = ∠MFE -> ME = MFAB ∥ MEN ∥ FGSo BN = NGBut MEN ∥ AB -> EN ⊥ AB => ENB = EBN = 45° -> EN = BNSo BN = NG = EN = rHence BG = 2r(-> denotes "implies")
Slight Correction:in the line"But MEN ∥ AB -> EN ⊥ AB => ENB = EBN = 45°"it should be read as " ∠BEN = ∠EBN = 45° "
Extend EF to cut AB at M; since M is the symmetrical of F about E and FG||AB, FG is tangent to the incircle and thus we are done.Best regards,Stan Fulger
Since FG is parallel to AB we can see that FE bisects < AFG. So E is the ex centre of Tr. FGC. Hence EG bisects right < BGF and so < EGB = 45. But < EBG = 45 and so BG= 2r where r is the in radius = altitude of isoceles Tr. BEGSumith PeirisMoratuwa Sri Lanka