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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1264.
To AntonioReferring to problem 1264, The line DF cut arc AC at 2 points G1 and G2.I understand that the concurrency of BH, EG and AC will be true for both points G per the problem statement.Do I understand it correctly ? Please clarify.Peter Tran
To Peter,Affirmative answer. The concurrency of BH, EG and AC will be true for both points G. Thanks.
Let AC, EG cut at X< DBF = < EBA = < EGA = < ECA = @ say ....(1)Let < ECB = < EAB = < EGB = € ...(2)Now < ADB = < € since < BCA = @ + €So GXBD is concyclicHence < XBG = < XDG = 90-@-€So < BHG = 90But < XHG = 90 so BXH are collinear points proving that AC, GE and and BH are concurrentSumith PeirisMoratuwaSri Lanka
Problem 1264Suppose EG intersects AC at point P, it suffices to show that the BP is perpendicular to AG.Suppose BK is the internal bisector of <ABC (K=means the arc AC),then KB perpendicular in ED.So arcAE=arcEC .Is <FBD=(<CAB+<ACB)/2=90-<ABC, <GDB=90-<FBD=<ABC/2 and <CDB=<BCA-<CBD=(<ACB-<BAC)/2. But <PGB=<EGB=<ECB=<ACB-<ACE=<ACB-(<CEB+<CDE)==(<ACB-<BAC)/2=<PDB. So the GPBD is cyclic.If BP intersect AG at H’ then <H’PG+<PGA==<GDB+<EGA=<ABC/2+<ECA=<KBC+<EKC=90.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
https://goo.gl/photos/NQrpJBvDM9aABBvB8Sumith and Apostolis already showed concurrency of BH, EG and AC. I will show that point G1 will have the same property as point G.Let EG1 meet AC at P and BP meet AG1 at H1We will show that BP⊥AG1 ( see sketch)Let u= ∠ (EBA)= ∠ (ECA)= ∠ (EG1A)= ∠ (DBC)And v=∠ (ECB)= ∠ (EG1B)In triangle BCD since ∠ (ACB)= u+ v and ∠ (CBD)= u=> ∠ (BDC)= vIn triangle CFD we have ∠ (FDC)= 90-u-v Since ∠ (PG1B)= ∠ (BDP)=v => BDG1P is cyclic and ∠ (PBG1)= ∠ (PDG1)= 90-u-vIn triangle BH1G1 ∠ (PBG1)+ ∠ (AG1B)= 90-u-v+u+v= 90So BP⊥AG1