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

Labels:
concurrent,
congruence,
equilateral,
external,
triangle

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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