Prove that two triangles are congruent ${ABC}$ and $A'B'C'$ are two triangles. Let $P$ be the midpoint of $BC$ and $P'$ the midpoint of B'C'. Also, $|AP| = |A'P'|$ and $|AC| = |A'C'|$ and $\angle CAB$ = $\angle C'A'B'$. $2|AP| > |AC|$ and $2|A'P'| > |A'C'|$. Prove that triangles $ABC$ and $A'B'C'$ are congruent. 
 A: Let $Q$ be the midpoint of $AC$, and likewise $Q'$.

It's enough to show that $\triangle APQ$ and $\triangle A'P'Q'$ are congruent (since $C$ is determined by $AQ$, and then $B$ is determined by $PC$).  The hypotheses yield that $|AP|=|A'P'|$ and $|AQ|=|A'Q'|$ and $\angle AQP=\angle A'Q'P'$ (since $\angle AQP = 180^\circ - \angle CQP = 180^\circ - \angle CAB$) and $|AP|>|AQ|$ and $|A'P'|>|A'Q'|$.  This is an instance of a version of SSA that actually works: SSA with the additional assumption that the side opposite the given angle is longer.  To prove this, superimpose $A'Q'$ and $AQ$, and angle $\angle A'Q'P'$ on $\angle AQP$:

Suppose for contradiction $P'$ doesn't coincide with $P$; WLOG, the points $P'\ast P\ast Q$ are in that order, as in the (impossible) figure.  Then $\angle AP'Q = \angle APP'$ (since $|AP|=|AP'|$; Euclid I:5), and $\angle APP' > \angle AQP'$ (as an exterior angle to a triangle; Euclid I:16), so $\angle AP'Q > \angle AQP'$, whence $|AQ|>|AP'|$ (as the sides opposite those angles; Euclid I:19), contrary to hypothesis.  (This argument is a variation on Euclid III:2.)
