Thursday, November 1, 2018

Geometry Problem 1396: Triangle with three rectangles on the sides, Vertices, Perpendicular lines, Concurrency

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Geometry Problem 1396: Triangle with three rectangles on the sides, Vertices, Perpendicular lines, Concurrency, Tutoring.
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1 comment:

  1. AnonymousNovember 2, 2018 at 3:28 AM

    Fie R = PB intersectat cu NA .
    Unghiul AEN = BAR ( unghiuri cu laturile perpendiculare).
    Unghiul AJN = CAR ( unghiuri cu laturile perpendiculare).
    Notam cu a, b, c unghiurile AEN ,BGP si CHQ .
    Atunci unghiurile AJN = CAR= A - a.
    Analog unghiurile PGB = CBR = b si PDB = ABR = B-b.
    Daca S = NA intersectat cu QC rezulta ca unghiul ACX = CHQ =c
    si unghiul BCX = CFQ = C – c.
    In triunghiul EAJ , sin( a )= AN/AE si sin J= sin (A –a ) = AN/AJ.
    De aici sin( a) / sin (A –a )= AJ/AE. (1)
    In mod analog obtinem sin( b)/sin(B-b) = BD/BG , (2)
    si sin ( c )/ sin (C- c) = CF/CH , (3).
    Daca inmultim relatiile (1) , (2) si (3) si cum AEDB, BGFC, ACHJ dreptunghiuri rezulta
    s sin( a) / sin (A –a )* sin( b)/sin(B-b)* sin ( c )/ sin (C- c)=1
    de unde conform teorema lui Ceva trigonometric rezulta AN, PB,QC concurente.
    Florin Popa Comanesti, Romania.

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