Saturday, November 14, 2009

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

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

 Problem 387. Triangle, Angle bisector, Perpendicular, Concyclic points.
See also:
Complete Problem 387
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

4 comments:

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

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  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

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  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

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

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete