The image below shows two triangles. I am convinced that this is a Side-Side-Angle situation, which is not enough to prove the two triangles congruent. Is this correct?
What is throwing me off is that the triangles share a common side.
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if $|NP|>|PM|$ then they are congruent, according to the definition of the SSA congruency. – Nov 26 '17 at 00:57
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Don't angles opposite congruent sides have to be congruent as well? – Michael McGovern Nov 26 '17 at 00:58
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@ThePirateBay, could you show me where SSA is a theorem about congruence? – Michael McGovern Nov 26 '17 at 01:00
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@MichaelMcGovern. I'm not sure what you're asking. "could you show me where SSA is a theorem about congruence?" - what do you mean? You can prove the examples from the link I've posted using trigonometry. – Nov 26 '17 at 01:05
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Even with a shared side, Side-Side-Angle —in and of itself— is not a congruence pattern. In your diagram, the relative positions of points $M$ and $O$ are ambiguous from the given information.
Blue
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Side-side-angle fails when there are two or no solutions. Your diagram presumably guarantees there is at least one solution. You get two solutions if you draw a circle around $P$ with radius $NP$ and it hits the ray $MN$ in two places. This is avoided if $NP \gt MP$ because the circle goes around behind $M$. If $NP$ is less than $MP$ but not too much less, $O$ could be higher on the page making $\angle NPO$ acute and the triangles are not congruent. If $NP$ is too much less than $MP$ the circle won't reach the ray and there is no solution.
Ross Millikan
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