Let ABC be any triangle. Equilateral triangles BCX, ACY, and BAZ are constructed such that none of these triangles overlaps triangle ABC.

a) Draw a triangle ABC and then sketch the remainder of the figure. It will help if triangle ABC is not isosceles (or equilateral).

b) Show that, regardless of choice of triangle ABC, we always have AX = BY = CZ.

*You must include a diagram for any complicated geometry problem that does not come with its own diagram. Diagrams are extremely important in geometry because they help your reader understand what is going on. Diagrams also help you to keep track of your work and possibly give you useful insight.*

Thanks so much!

AnonymousConfusedGuy Feb 6, 2018

#3**+2 **

Using your diagram, I can prove that AX = BY

Note that we have two triangles, ACX and BCY

AC = CY = "b"

And

XC = BC = "a"

And angle ACX = angle YCB

So.... by SAS .... triangle ACX is congruent to triangle YCB

So... AX = BY

But...I have not figured out how to prove that CZ is also equal to these

Maybe someone can provide the final step ???

CPhill Feb 7, 2018

#4**+3 **

Huh I guess this was a harder problem than I thought! Sorry for the impatience, it was just due yesterday so I got really antsy. Thanks for the help everyone!

AnonymousConfusedGuy
Feb 8, 2018

#5**+1 **

Thanks Chris,

The bit that follows is exactly the same as Chris said and then i have continued the proof

I am going to call the angles in the original triangle

All angles in the 3 other triangles are all 60 degrees because they are equilateral triangles.

Consider

\(\triangle ACX\;\; and \;\; \triangle YCB \\ AC=b, \;\;CX=a\;\; and \;\; \angle ACX=\angle C+60^\circ\\ YC=b, \;\;CB=a\;\; and \;\; \angle YCB=\angle C+60^\circ\\ \therefore \triangle ACX\cong \triangle YCB\qquad \text{(2 sides and included angle test)} \\ \therefore AX=BY \qquad \text{Corresponding sides in congruent triangles.}\)

\(\text{Now consider} \triangle ABY\;\; and \;\; \triangle AZC \\ AB=AZ=c\\ AY=AC=b\\ \angle YAB=\angle CAZ=\angle A+60^\circ\\ \therefore \triangle ABY\cong \triangle AZC\qquad \text{(2 sides and included angle test)} \\ \therefore BY=ZC \qquad \text{Corresponding sides in congruent triangles.}\\ \)

\(\therefore AX=BY=ZC\\~\\ QED\)

Melody Feb 8, 2018

#8**+2 **

OH! Thanks so much! I can use this actually for another problem that stumped me!

AnonymousConfusedGuy
Feb 8, 2018