SSA

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, $\mathrm{△}ABC$ and $\mathrm{△}PQR$. $\mathrm{△}ABC\cong \mathrm{?}\mathrm{△}PQR$ if $\overline{AB}\cong \overline{PQ}$, $\overline{BC}\cong \overline{QR}$, and either $\mathrm{\angle }BAC\cong \mathrm{\angle }QPR$ or $\mathrm{\angle }BCA\cong \mathrm{\angle }QRP$.

Since this method does not prove congruence, it is more useful for disproving it. If the SSA method is attempted between $\mathrm{△}ABC$ and $\mathrm{△}PQR$ and fails for every $ABC$,$BCA$, and $CBA$ against every $PQR$,$QRP$, and $RPQ$, then $\mathrm{△}ABC\cong ̸\mathrm{△}PQR$.

Suppose $\mathrm{△}ABC$ and $\mathrm{△}PQR$ the SSA test. The specific case where SSA fails, known as the ambiguous case, occurs if the congruent angles, $\mathrm{\angle }BAC$ and $\mathrm{\angle }QPR$, are acute. Let us illustrate this.

Suppose we have a right triangle, $\mathrm{△}XYZ$, with right angle $\mathrm{\angle }XZY$. 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 $\mathrm{\angle }XZY$ is right, this makes $\mathrm{\angle }PZY$ 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, $\mathrm{\angle }YXP$ is in fact the same angle as $\mathrm{\angle }YXQ$, thus $\mathrm{\angle }YXP\cong \mathrm{\angle }YXQ$. Since $\overline{XY}\cong \overline{XY}$, $\mathrm{△}XYP$ and $\mathrm{△}XYQ$ clearly meet the SSA test, and yet, just as clearly, are not congruent. This results from $\mathrm{\angle }YXZ$ being acute. This example also reveals the exception to the ambiguous case, namely $\mathrm{△}XYZ$. 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 $\mathrm{\angle }XZY$ 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.

Title	SSA
Canonical name	SSA
Date of creation	2013-03-22 12:28:53
Last modified on	2013-03-22 12:28:53
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	6
Author	mathcam (2727)
Entry type	Definition
Classification	msc 51M99