SOLUTION: 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

Algebra ->  Geometry-proofs -> SOLUTION: 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       Log On


   

Question 1025587: 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
Use much BASIC proofs as possible.
Answer by ikleyn(52835) About Me  (Show Source):
You can put this solution on YOUR website!
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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
Use much BASIC proofs as possible.
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Solution 1 
Let us prove that  |AX| = |BY|.  For other combinations of segments the proofs will be the same.

From the triangle ACX, according to the cosines law

abs%28AX%29%5E2 = abs%28AC%29%5E2+%2B+abs%28CX%29%5E2+-+2%2Aabs%28AC%29%2Aabs%28CX%29%2Acos%28ACX%29.   (1)   ( argument ACX under cos is the angle ACX ).

Notice that LACX = LC + 60°,  where LC is the angle C of the original triangle ABC.

From the triangle BCY, according to the cosines law

abs%28BY%29%5E2 = abs%28CY%29%5E2+%2B+abs%28BC%29%5E2+-+2%2Aabs%28CY%29%2Aabs%28BC%29%2Acos%28BCY%29.   (2)   ( argument BCY under cos is the angle BCY ).

Notice that LBCY = LC + 60°,  where LC  is again the angle C of the original triangle ABC.

Now,  |AC| = |CY|,  |CX| = |BC|  and  LACX = LC + 60° = LBCY. 

Therefore, right sides of (1) and (2) are equal.
Hence, their left sides are equal.
It implies that  |AX| = |BY|.

The statement is proved.

If you make a sketch, the proof will be crystally clear to you.

Solution 2     (even more straightforward) 
Again, let us prove that  |AX| = |BY|. 

Consider triangles  ACX  and  BCY.

They have one pair of congruent sides CX  and  BC, and the other pair of congruent sides  AC  and  CY.

They also have congruent angles ACX and BCY that are concluded between the corresponding sides of these pairs.

Indeed, each angle ACX and BCT is the angle C of the original triangle ABC plus the angle of 60°.

So, the triangles  ACX  and  BCY  are congruent according with SAS-test for triangles congruency.

Thus  |AX| = |BY|, what has to be proved.


For other combinations of segments the proofs are very similar.

Curious observations / corollaries 
After the second proof, it becomes clear that if you rotate the triangle  ACX  around the vertex C in 60°, you will get the triangle  YCB. 

It implies that the (directed) segments  AX  and  BY  make the angle of 120° (taking into account the directions of these segments).


So, the tree segments AX,  BY  and  CZ  make the angles of 120° between them.