# 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, $\mathrm{\u25b3}ABC$ and $\mathrm{\u25b3}PQR$. $\mathrm{\u25b3}ABC\cong \mathrm{?}\mathrm{\u25b3}PQR$ 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 $\mathrm{\u25b3}ABC$ and $\mathrm{\u25b3}PQR$ and fails for every $ABC$,$BCA$, and $CBA$ against every $PQR$,$QRP$, and $RPQ$, then $\mathrm{\u25b3}ABC\cong \u0338\mathrm{\u25b3}PQR$.

Suppose $\mathrm{\u25b3}ABC$ and $\mathrm{\u25b3}PQR$ 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, $\mathrm{\u25b3}XYZ$, 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 $\overline{XY}\cong \overline{XY}$, $\mathrm{\u25b3}XYP$ and $\mathrm{\u25b3}XYQ$ 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 $\mathrm{\u25b3}XYZ$. 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 |