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I am trying to understand a proof of Pythagorean's Theorem.

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I am trying to see how all three triangles are similar. So far I have:

(1) $\triangle ABC$ and $\triangle DBA$ have side $AB$ in common and also $\angle ABC$. What is the other side?

(2) $\triangle ABC$ and $\triangle DAC$ have $\angle BCA=\angle DCA$ and side $AC$ in common. What is the other side?

(3) $\triangle DBA$ and $\triangle DAC$ have side $AD$ in common...what else?

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(1) $\triangle ABC$ and $\triangle DBA$ have side $AB$ in common and also $\angle ABC$. What is the other side?

The fact that these two triangles have a side in common is a red herring! The line segment $AB$ just happens to be a side of both triangles, but this is irrelevant to the fact that the triangles are similar.

If you look at $\triangle ABC$ and $\triangle DBA$, you will see that their shorter legs are $AB$ and $DB$ (respectively), their longer legs are $AC$ and $DA$ (respectively), and their hypotenuses are $BC$ and $BA$ (respectively). If you look at each of these pairs of line segments, and calculate the ratio of each pair, you'll find that the three ratios are all equal.

The corresponding angles are $\angle ABC$ with $\angle DBA$ (the wider acute angles); $\angle BCA$ with $\angle BAD$ (the narrower acute angles); and $\angle CAB$ with $\angle ADB$ (the right angles). For each of these three pairs of angles, both of the angles in the pair are congruent.

Does this help explain things?

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  • $\begingroup$ yes, thank you! $\endgroup$ – rover2 Sep 20 '18 at 13:31
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Maybe it's easier to look at the angles.

For instance $\Delta ABC\sim\Delta DBA$ since $\angle BAC = \angle BDA$ and both triangles have a right angle. The same argument can be used to show $\Delta ABC ~\sim\Delta DAC$. From the first two similarities, the last similarity follows automatically.

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