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Proposed ProblemSee also: Complete Problem 31, Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
As proven in Problem 30, BG=2r. Similarly BH=2r or BH=BG=2r. Now if B be the origin, BA the x-axis and BC the y-axis then E is (r,r) while HG is: x + y = 2r since H is (2r,0)& G:(0,2r). Clearly E satisfies the equation of HG; hemce E lies on HG or H, E & G are collinear.Ajit: firstname.lastname@example.org
In problem 30, we see that ang(BEG)=ang(BEQ)+ang(QEG)=45+45=90 andang(BEH)=ang(BEP)+ang(PEH)=45+45=90 Hence H, E and G are collinear.
expand DH and GF at I.because angle (BHD) = angle (BGF) = 90 so BGHI is a QUADRILATERAL that lies on circle E.Angle (2B)=angle(E).so anle (HBG)=180 that means it is collinear
if ang(EAD)=a and ang(EAC)=45-a we can easily find that ang(EFD)=ang(EFG)=90-a which means EFid the bisector of ang(GFD) and EC the bisector of ang(GCF)so E is the excenter of FGC.Then ang(BGE)=ang(EGF)=45 and similarly ang(EHD)=45 so ang(BGE)=ang(EHD) Because BG//HD H,E,G are collinear
Kindly refer my solution to Problem 30. We showed that Tr. BEG is an isoceles right Tr. It can similarly be shown that Tr. BEH is also a isoceles right Tr. Hence < HEB + < GEB = 90 + 90 = 180 proving that H, E,G are collinearSumith PeirisMoratuwaSri Lanka
If HD, GF meet at M, then HBGM is a square with side 2r and centre EFurther it can be shown that < DEF = 45 and that E is ex centre of Tr. DMF which is a Tr. similar to the original Tr. ABCSumith PeirisMoratuwaSri Lanka