MHB Square metric not satisfying the SAS postulate

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The square metric distance in R^2 is defined as D((x1, y1), (x2, y2))= max{|x2 − x1|, |y2 − y1|}. An example is provided using two triangles: the first triangle with vertices (0,0), (1,0), and (0,1) has all sides of metric length 1. When this triangle is rotated by 45 degrees to form a second triangle with vertices (0,0), (1,1), and (1,-1), the side lengths change, resulting in lengths of 1, 1, and 2. This demonstrates that the square metric does not satisfy the SAS postulate, as congruency requires all sides to be equal. Thus, the square metric fails to uphold the SAS postulate in this context.
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I'm not sure on how to do this problem. Can someone please help and explain? Thank you!

Recall (Exercise 3.2.8) that the square metric distance between two points (x1, y1) and
(x2, y2) in R^2 is given by D((x1, y1), (x2, y2))= max{|x2 − x1|, |y2 − y1|}. Show by
example that R^2 with the square metric and the usual angle measurement function does
not satisfy the SAS Postulate.
 
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pholee95 said:
I'm not sure on how to do this problem. Can someone please help and explain? Thank you!

Recall (Exercise 3.2.8) that the square metric distance between two points (x1, y1) and
(x2, y2) in R^2 is given by D((x1, y1), (x2, y2))= max{|x2 − x1|, |y2 − y1|}. Show by
example that R^2 with the square metric and the usual angle measurement function does
not satisfy the SAS Postulate.

Hi pholee95! Welcome to MHB! ;)

Let's pick a simple triangle, like a rectangular one with points (0,0), (1,0), (0,1).
Now let's rotate it by 45 degrees, keeping the angles the same, and keeping the lengths of the side according to the metric the same.
Then we'll get the triangle with points (0,0), (1,1), (1,-1).
According to the SAS postulate it should then be congruent.
But congruency requires that the lengths of all sides are the same. Is the (metric) length of the 3rd side the same?
 
I like Serena said:
Hi pholee95! Welcome to MHB! ;)

Let's pick a simple triangle, like a rectangular one with points (0,0), (1,0), (0,1).
Now let's rotate it by 45 degrees, keeping the angles the same, and keeping the lengths of the side according to the metric the same.
Then we'll get the triangle with points (0,0), (1,1), (1,-1).
According to the SAS postulate it should then be congruent.
But congruency requires that the lengths of all sides are the same. Is the (metric) length of the 3rd side the same?

No it won't be the same. Right?
 
pholee95 said:
No it won't be the same. Right?

Correct. All sides of the first triangle have metric length 1.
But the second triangle has metric lengths 1, 1, and 2.
 

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