# SSA can prove of the congruence of two triangles?

Hey folks! I recently made a project about the congruence of triangles in an advanced way. The one of the topic is SSA (Side-side-angle). Teachers always say that SSA cannot prove two triangles are congruent (the explanation is here), but I want to tell you guys about SSA can prove the congruence of two triangles with some conditions. Let's check this out!

At first, let a triangle be $\Delta ABC$ such that we only know about the length of $AB, BC$ and the size of $\angle A$, by the sine formula, we can show that

$\dfrac { BC }{ \sin { \angle A } } =\dfrac { AB }{ \sin { \angle C } } \rightarrow \sin { \angle C } = \dfrac { AB\sin { \angle A } }{ BC }$

From here there are two cases:

Case $1$, $\sin { \angle C }=1$

There are only $1$ possibility of $\angle C$, which is $90\degree$. Then, we know the size of $\angle A, \angle C$ and the length of $AB$, so we can show that there are only $1$ possible triangle by applying $AAS$. Therefore, we can conclude that when $\sin { \angle C }=\dfrac { AB\sin { \angle A } }{ BC }=1$, which means $\dfrac { BC }{ AB }=\sin { \angle A }$, $SSA$ is true.

Case $2$, $0<\sin { \angle C }<1$

Then, there are $2$ possibilities of $\angle C$, $\sin ^{ -1 }{ \left(\dfrac { AB\sin { \angle A } }{ BC }\right) }$ and $180 \degree-\sin ^{ -1 }{ \left(\dfrac { AB\sin { \angle A } }{ BC }\right) }$. Hence, we need to consider $\angle A$ to find out more conditions of proving $SSA$. By the angle sum of triangle, $\angle C=180\degree-\angle A-\angle B<180\degree-\angle A$. We will consider two cases about $\angle A$

Case $A$, $\angle A \ge 90\degree$

$\angle C<180\degree-\angle A\le 90\degree$. Then, $\angle C$ can only be $\sin ^{ -1 }{ \left(\dfrac { AB\sin { \angle A } }{ BC }\right) }$, so there are only $1$ possible triangle similarly. Therefore, we can conclude that when $\angle A \ge 90\degree$, $SSA$ is true.

Do you notice that? When $\angle A = 90\degree$, it is the same case as RHS (Right angle-hypotenuse-side) or HL (Hypotenuse-leg)

Case $B$, $\angle A < 90\degree$

$\angle C<180\degree-\angle A$. It is true when $\angle C=\sin ^{ -1 }{ \left(\dfrac { AB\sin { \angle A } }{ BC }\right) }$. Because of we want to have $1$ possibility of $\angle C$, we need to ignore the other possibility of $\angle C$, which we get the inequality below:

$90\degree>180\degree-\sin ^{ -1 }{ \left( \dfrac { AB\sin { \angle A } }{ BC } \right) } \ge 180\degree-\angle A\\ \\ \sin { 90\degree } >\sin { \left( 180\degree-\sin ^{ -1 }{ \left( \dfrac { AB\sin { \angle A } }{ BC } \right) } \right) } \ge \sin { \left( 180\degree-\angle A \right) } \\ \\ \dfrac { AB\sin { \angle A } }{ BC } \ge \sin { \angle A } \\ \\ AB\ge BC$

Therefore, we can conclude that when $AB\ge BC$, $SSA$ is true.

To sum up, there is a triangle $\Delta ABC$ given the length of $AB, BC$ and the size of $\angle A$, if this triangle has one or more of the conditions below, then $SSA$ is true:

• $\dfrac { BC }{ AB }=\sin { \angle A }$
• $\angle A \ge 90\degree$
• $AB\ge BC$

Note by Isaac Yiu Math Studio
6 months, 4 weeks ago

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There are many cases it doesn't prove congruency.

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How $\ \ \ 90 \gt 180 -\sin ^{-1} (\frac {c}{a} \sin A) \geq 180 - A \ \$ is true ?

As, $\ \ A < 90 \ \ So \ \ 180 - A \gt 90$

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