Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 574.Zoom at: Geometry Problem 574
Let FM intersection CD be GDenote the "acute" angles at A, B, C, D, E, F, Gby the corresponding small letters a, b, c, d, e, f, g.b = d = f + c f = b – c d = g + (f/2) , g = d – (f/2) = b – [(b-c)/2] = (b + c)/2e + b + c = Pi,(e/2) = (Pi/2) – [(b+c)/2] = (Pi/2) – g (e/2) + g = Pi/2 in triangle EHGSo angle at H = 90 degrees
http://img137.imageshack.us/img137/1631/problem574.pngEH cut AD at K and BC at L ( see attached picture)Note that Triangle KED similar to tri. LEB (case AA)So angle(LKF)=angle(KLF) and triangle KLF is isoscelesAnd angle bisector FH of isosceles triangle KLF perpendicular to the base KL at HPeter Tran
name P intersection of AC, DB extended▲AEP ~ ▲DEP ( CAB = CDB, AEP = DEP )=> APH = DPH => PH bisector from 572PH perpendicular to FH
Problem #574. To c.t.e.o: What is the reason for your statement angle AEP = angle DEP?
To AntonioHEP bisector of CEB, AEC = DEBAEC + 1/2 CEB = DEB + 1/2 CEB
Problem #574. To c.t.e.o: Thanks for your comment. What is the reason for your statement HEP is the bisector of CEB (why are points H, E, and P collinear?). Thanks.
To Antonio,Thanks. H, E, P are never collinearHEP is the same bisector for vertcales anglesWe have to prove bisector of P // to bisector of CBEAC, DB meet at P1, bisector of CEB meet DP1 at Pbisector of P1 meet CD at E1In ▲P1CE1 1/2 P1 + P1CE1 + CE1P1 = 180In E1EPP1 180- CE1P1 + E1EP + E1EP + PBE + 1/2P1=360from both => 2CE1P1 = 2E1EP => CE1P1 = E1EPas corresponding angle => P1E1//PE
Problem #574, To c.t.e.o: Thanks.
Let G cut BC at X and AD at Y. Then < A = < C. Hence Tr. FXY is isoceles and the result follows.Sumith PeirisMoratuwaSri Lanka