Proposed Problem

Click the figure below to see the complete problem 387.

See also:

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

Labels:
angle bisector,
concyclic,
perpendicular,
triangle

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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