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# Isosceles & Equilateral Triangles, Perimeter, and Area

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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 \(AX=BY=CZ\).

benjamingu22  Aug 23, 2017
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#1
+91040
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We need someone better than me at geometry proofs here

Any takers  ??

Thios is the diagram.

The aim is to show that the 3 red lines are equal in length.

Melody  Aug 25, 2017
#2
+78729
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This isn't as difficult as it first appears....although I have to admit, it threw me for a bit...!!!

Note that we have triangles ZAC and BAY

And two sides of ZAC - namely, b and c - are equal to two sides of BAY

And angle ZAB + angle CAB  = angle YAC + angle CAB

Thus....by SAS......triangle ZAC is congruent to triangle BAY....therefore ZC  = BY

Likewise...we have two other triangles, BCY and XCA

And two sides of BCY - namely, a and b - are equal to to sides of XCA

And angle XCB + angle ACB  = angle YCA + angle ACB

Thus.....by SAS......triangle BCY  is congruent to triangle XCA

Thus BY  = XA  = AX

But BY was also proved equal to ZC = CZ

Therefore....  AX = BY = CZ

P.S. - Melody should get some points for providing the diagram....without that, I wouldn't have understood the layout....!!!!

CPhill  Aug 25, 2017
edited by CPhill  Aug 25, 2017
#3
+91040
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Thanks Chris, much appreciated.

I still have to think about it.

Maybe it is still too early in the morning for me. I have to have some excuse.

Melody  Aug 25, 2017

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