4.2: Angle-Side-Angle Condition

Theorem $\PageIndex{1}$ ASA condition

Assume that

$AB = A'B'$, $\measuredangle ABC = \pm \measuredangle A'B'C'$, $\measuredangle CAB = \pm \measuredangle C'A'B'$

and $\triangle A'B'C'$ is nondegenerate. Then

$\triangle ABC \cong \triangle A'B'C'$.

Note that for degenerate triangles the statement does not hold. For example, consider one triangle with sides 1, 4, 5 and the other with sides 2, 3, 5.

Proof

According to Theorem 3.3.1, either

$$\begin{array} {l} {\meausredangle ABC = \measuredangle A'B'C',} \\ {\measuredangle CAB = \measuredangle C'A'B'} \end{array}$$

or

$$\begin{array} {l} {\meausredangle ABC = -\measuredangle A'B'C',} \\ {\measuredangle CAB = -\measuredangle C'A'B'.} \end{array}$$

Further we assume that 4.2.1 holds; the case 4.2.2 is analogous.

Let $C''$ be the point on the half-line $[A'C')$ such that $A'C'' = AC$.

By Axiom IV, $\triangle A'B'C'' \cong \triangle ABC$. Applying Axiom IV again, we get that

$\measuredangle A'B'C'' = \measuredangle ABC = \measuredangle A'B'C'$.

By Axiom IIIa, $[B'C') = [BC'')$. Hence $C''$ lies on $(B'C')$ as well as on $(A'C')$.

Since $\triangle A'B'C'$ is not degenerate, $(A'C')$ is distinct from $(B'C')$. Applying Axiom II, we get that $C'' = C'$.

Therefore, $\triangle A'B'C' = \triangle A'B'C'' \cong \triangle ABC$.