Proposed Problem
Click the figure below to see the complete problem 387.
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Complete Problem 387
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Saturday, November 14, 2009
Problem 387. Triangle, Angle bisector, Perpendicular, Concyclic points
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For treangles BFH, BFE, BFG, DFD Phifagor theorem BF^2=BH^2+FH^2 and etc. Then BF is diametr of this circle.
ReplyDeletejoin B,F.sinve angleBHF+angleBGF=180 so B,H,F,S lie on same circle.simillarly the points B,E,F,D also concyclic.and angle BHF=BEF=90 so the points B,H,E,F are concylic and the points B,G,D,F are concyclic hence the points B,H,E,F,D,G discribe the same circle
ReplyDeleteThere are 4 right angles at E, F, G and H and all these lie on a circle with BF as diameter
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka
Dear Antonio - AF and CF need not be angle bisectors for this result to hold. Any point F inside the Triangle ABC would just as do
DeleteDear Sumith,
DeleteYou are correct; the condition of bisectors is not necessary. Perhaps I was thinking about other discoveries that have yet to be explored. Thank you.
Antonio
Reference my proof for Problem 385, < BED = < EBG = < DGC since BEDG is cyclic and the result follows
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka
<BHF+<BCF=180
ReplyDeleteBHFC is cyclic
<BEF+<BDF=180
BEFD is cyclic
<BEF=<BHF
BHFE is cyclic
Combining the above 3 results, BGDFEH are cyclic