SSA 2002-02-25 15:37:28 2003-10-16 18:58:05 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \PMlinkescapeword{meet} \emph{SSA} is a method for determining whether two triangles are congruent by comparing two sides and a non-inclusive angle. However, unlike SAS, SSS, ASA, and SAA, this does not prove congruence in all cases. Suppose we have two triangles, $\triangle ABC$ and $\triangle PQR$. $\triangle ABC \cong? \triangle PQR$ if $\overline{AB} \cong \overline{PQ}$, $\overline{BC} \cong \overline{QR}$, and either $\angle BAC \cong \angle QPR$ or $\angle BCA \cong \angle QRP$. Since this method does not prove congruence, it is more useful for disproving it. If the SSA method is attempted between $\triangle ABC$ and $\triangle PQR$ and fails for every $ABC$,$BCA$, and $CBA$ against every $PQR$,$QRP$, and $RPQ$, then $\triangle ABC \not\cong \triangle PQR$. Suppose $\triangle ABC$ and $\triangle PQR$ \PMlinkescapetext{meet} the SSA test. The specific case where SSA fails, known as the ambiguous case, occurs if the congruent angles, $\angle BAC$ and $\angle QPR$, are acute. Let us illustrate this. Suppose we have a right triangle, $\triangle XYZ$, with right angle $\angle XZY$. Let $P$ and $Q$ be two points on $\overleftrightarrow{XZ}$ equidistant from $Z$ such that $P$ is between $X$ and $Z$ and $Q$ is not. Since $\angle XZY$ is right, this makes $\angle PZY$ right, and $P$,$Q$ are equidistant from $Z$, thus $\overleftrightarrow {YZ}$ bisects $P$ and $Q$, and as such, every point on that line is equidistant from $P$ and $Q$. From this, we know $Y$ is equidistant from $P$ and $Q$, thus $\overline{YP} \cong \overline{YQ}$. Further, $\angle YXP$ is in fact the same angle as $\angle YXQ$, thus $\angle YXP \cong \angle YXQ$. Since $\overline{XY} \cong \overline{XY}$, $\triangle XYP$ and $\triangle XYQ$ clearly meet the SSA test, and yet, just as clearly, are not congruent. This results from $\angle YXZ$ being acute. This example also reveals the exception to the ambiguous case, namely $\triangle XYZ$. If $R$ is a point on $\overleftrightarrow{XZ}$ such that $\overline{YR} \cong \overline{YZ}$, then $R \cong Z$. Proving this exception amounts to determining that $\angle XZY$ is right, in which case the congruency could be proven instead with SAA. However, if the congruent angles are not acute, i.e., they are either right or obtuse, then SSA is definitive.