Proposed Problem
Click the figure below to see the complete problem 404 about External Equilateral triangles, Congruent and Concurrent Lines.
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Complete Problem 404
Level: High School, SAT Prep, College geometry
Thursday, December 17, 2009
Problem 404. External Equilateral triangles, Congruent and Concurrent Lines
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half of solution
ReplyDeleteTr 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'
tri ABA'=~tri C'BC (proved)
ReplyDelete=> 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.
Let AA’ and CC’ meet at X
ReplyDeleteJoin BX, B’X
Tr.s BCC’ ≡ BAA’ (SAS) so AA’ = CC’. Similarly each of these are = to BB’.
Also < BAA’ = < BCC’, so AXBC’ is concyclic
Therefore < AXC’ = BXC’ = BXA’ = 60 .….(1) (being respectively equal to the
three 60 degree angles of Equilateral Tr. ABC’)
Now AXC’ = 60 = AB’C, hence AB’CX is concyclic and so < AXB’ = < ACB’ = 60
….(2)
But AXB = < AXC’ + BXC’ = 60 + 60 =120….(3) {from (1)}
From (2) and (3) < B’XB = 120 + 60 = 180, so BXB’ are collinear points and
so lines AA’, BB’ and CC’ are concurrent.
Sumith Peiris
Moratuwa
Sri Lanka