Question 1199620
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What you're observing isn't a coincidence.


Consider a right triangle ABC such that segment BC is the base, and point A is directly above B.


Copy triangle ABC and shift it over exactly BC units to the right. This will have points B' and C line up perfectly.


Now connect points A and A' to form a segment. Refer to figure 1 shown below.
This segment is the same length as side BC. We have rectangle ABCA', which is why AA' is congruent to BC.


The last thing to do is nudge point A some amount to the right (see figure 2). The amount doesn't matter. Let's say it's 1 unit to the right. This same nudge is applied to A' as well. Both points move the same amount to the right, which means segment AA' doesn't change in length. The segment is simply translated that amount to the right. Hence AA' is still congruent to BC.


Example Diagrams
*[illustration Screenshot_222.png]
and
*[illustration Screenshot_223ccc.png]

Another way to look at it: The quadrilateral ABCA' in either diagram is a parallelogram. Meaning we have two sets of parallel opposite sides. One such property is that the opposite sides are congruent. It's similar to a rectangle but the sides can shear and skew.


If you want to prove this using triangle congruences, then focus on triangles ABC and B'A'A. The order of the lettering is important so we can determine how the letters pair up together.


Side note:
GeoGebra was used to make the diagrams shown. It's a tool similar to Desmos.


Yet another way to look at it:
If you shifted triangle ABC exactly BC units to the right, then point A moves the same distance. It must move this distance so it stays attached to the triangle we're moving. Otherwise, it'll transform the second triangle into something that isn't congruent to the first triangle.
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