Saturday, November 14, 2009

Problem 387. Triangle, Angle bisector, Perpendicular, Concyclic points

Proposed Problem
Click the figure below to see the complete problem 387.

Complete Problem 387
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

1. For treangles BFH, BFE, BFG, DFD Phifagor theorem BF^2=BH^2+FH^2 and etc. Then BF is diametr of this circle.

2. join 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

3. There are 4 right angles at E, F, G and H and all these lie on a circle with BF as diameter

Sumith Peiris
Moratuwa
Sri Lanka

1. 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

2. Dear Sumith,
You 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

4. Reference my proof for Problem 385, < BED = < EBG = < DGC since BEDG is cyclic and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

5. <BHF+<BCF=180
BHFC is cyclic

<BEF+<BDF=180
BEFD is cyclic

<BEF=<BHF
BHFE is cyclic

Combining the above 3 results, BGDFEH are cyclic