Thursday, December 17, 2009

Problem 404. External Equilateral triangles, Congruent and Concurrent Lines

Proposed Problem
Click the figure below to see the complete problem 404 about External Equilateral triangles, Congruent and Concurrent Lines.

Problem 404. External Equilateral triangles, Congruent and Concurrent Lines.
See also:
Complete Problem 404
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 5:55 PM
Labels: , , , ,

2 comments:

  1. half of solution

    Tr AA'B and tr CC'B have

    1) BA' = BC
    2) AB = BC'
    3) an C'BC = an A'BA ( 60 + an B )

    => AA' = CC'

    Tr BB'C and AA'C at the same way give

    => BB' = AA'

    ReplyDelete

  2. tri ABA'=~tri C'BC (proved)
    => ang BAA'=ang BC'C.
    let AA' and CC' meet at P.
    ang BAP = ang BC'P.
    =>BC'AP is cyclic (angles in the same segment).
    =>ang BPA = 120 deg.(opp. angles of a cyclic quad. are supplementary)
    similarly ang BPC = 120 deg.(Don't consider B,P,B').
    =>ang APC = 120 deg.(angles around a point).
    =>APCB' is cyclic(since angAPC + angAB'C = 180 deg)
    =>ang APB' = ang ACB' = 60 deg.
    ang APB' = 60 deg , and
    ang APB = 120 deg.
    => B,P,B' are collinear.(since the angle between them is 180 deg).
    => AA', BB', CC' are concurrent.

    ReplyDelete

Subscribe to: Post Comments (Atom)