## Monday, February 2, 2009

### Elearn Geometry Problem 240: Triangle with Equilateral triangles, Parallelogram See complete Problem 240 at:
gogeometry.com/problem/p240_triangle_equilateral_parallelogram.htm

Level: High School, SAT Prep, College geometry

1. ang C'AC = 60 - A =>
B'AC' = A
A'CB' = C

extend BA' to P ( P on AB'), BC' to G, ( G on B'C )
BPB' = B - 60 + 60 + A, BGB' = B - 60 + 60 + C =>

BPB' = A + B
BGB' = B + C

▲ABC and ▲AC'B' are congruent
1) AB = AC'
2) BAC = C'AB' = A
3) AC = AB'
=>
AC'B' = B, AB'C' = C
=>
BPB' + AB'C' = (A + B) + C = 180°
=>

BP//B'C'

BGB' + GB'A' = (B + C) + A = 180°
=>

BC'//A'B'
------------------------------------------

2. 1. Triangles AA’B’ and ABC are congruent ( AB’=AC , AC’=AB, angle B’AC’= angle CAB ) so BC’=BC=BA’
2. Triangles CA’B’ and CBA are congruent ( CB’=CA, CA’=CB, angle B’CA’= angle ACB)
so B’A’=BA=BC’
3. Quadrilateral BA’B’C’ have opposite sides congruent so BA’B’C’ is a parallelogram.

3. See the drawing

ΔCAB’ and ΔC’AB are equilateral
=> ∠CAB’=∠C’AB=Π/3
∠BAB’=a+∠CAB’=∠BAC’+a’
=> a=a’
AC’=AB and AB’=AC and a=a’=> ΔBAC and ΔB’AC’ are congruent (SAS)
=>B’C’=BC
But BC=BA’=>B’C’=BA’
In the same way : ∠BCA’=∠B’CA=Π/3
∠BCA=c= Π/3 - ∠ACA’
∠B’CA’=c’= Π/3 - ∠ACA’ =>c=c’
CB=CA’ and CA=CB’ and c=c’=> ΔBCA and ΔA’CB’ are congruent (SAS)
=>AB=A’B’
But AB=BC’ => A’B’=BC’

B’C’=BA’ and A’B’=BC’ => BA’B’C’ is a parallelogram

4. It is not difficult to see that < A'CB' = C and hence Tr.s A'B'C & ABC are congruent whence A'B' = BC'

By similar reasoning A'B = B'C'

In quadrilateral A'B'C'B the opposite sides are thus equal and the result follows.

Sumith Peiris
Moratuwa
Sri Lanka