## 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. Complete Problem 404
Level: High School, SAT Prep, College geometry

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'

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.

3. Let AA’ and CC’ meet at X
Join 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