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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.This entry contributed by Ajit Athle.Click the figure below to see the problem 751 details.
http://img204.imageshack.us/img204/4091/problem751.pngExtend BG to cut CD at H .Extend CG to cut BA at P ( see sketch)Let AB=b , BC=a, ED=BF=x∆ (BCH) similar to ∆ ( EDH) so HC/a=HD/x=(HC-HD)/(a-x)= b/(a-x)So HD=bx/(a-x) and CH=ab/(a-x)∆ (FBG) similar to ∆ (DGH) and ∆ (PBG) similar to ∆ (CHG)So BF/DH=BG/GH=BP/CH=(a-x)/bAnd BP=CH (a-x)/b = ab/(a-x). (a-x)/b= aSo BP=BC=a and triangle CBP is isosceles ∠ (BCP)= ∠ (BPC)= ∠ (PCD)….(Alternate angles)CG bisect ∠ (BCD)
Extend BE and CD to meet at H.BC/CH = ED/DH = BF/DH =BG/GHHence CG bisects angle BCH
Proof:Through G construct PQ // AD so that P on AB and Q on DC Through G construct RS // AB so that R on AD and S on BC Triange GPB and GRE are similarPB : PG = RG : RETriangle GPF and DRG are similarPF : PG = RG : RDPF : PB = RE : RD(PB – BF) : PB = (RD – DE) : RDBF : PB = DE : RDBF = DEPB = RDGSCQ is a rhombus CG is the bisector of angle BCD
Dear Pravin,Sorry that I don't understand your proof.Why a pair of triangles with two sides ratios equal them their are similar?Kent
Triangles BCH and EDH are similar-> first two equalities;BF=ED -> 2 and 3rd equality;BFG and GDH are similar-> last two equalities. And BC/CH=BG/GH is property of bisector,hence GC is bisector of angle C. I solved problem completely indetical:)