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 ΔABC\Delta ABC such that we only know about the length of AB,BCAB, BC and the size of A\angle A, by the sine formula, we can show that

BCsinA=ABsinCsinC=ABsinABC\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 11, sinC=1\sin { \angle C }=1

There are only 11 possibility of C\angle C, which is 90°90\degree. Then, we know the size of A,C\angle A, \angle C and the length of ABAB, so we can show that there are only 11 possible triangle by applying AASAAS. Therefore, we can conclude that when sinC=ABsinABC=1\sin { \angle C }=\dfrac { AB\sin { \angle A } }{ BC }=1, which means BCAB=sinA\dfrac { BC }{ AB }=\sin { \angle A }, SSASSA is true.

Case 22, 0<sinC<10<\sin { \angle C }<1

Then, there are 22 possibilities of C\angle C, sin1(ABsinABC)\sin ^{ -1 }{ \left(\dfrac { AB\sin { \angle A } }{ BC }\right) } and 180°sin1(ABsinABC)180 \degree-\sin ^{ -1 }{ \left(\dfrac { AB\sin { \angle A } }{ BC }\right) }. Hence, we need to consider A\angle A to find out more conditions of proving SSASSA. By the angle sum of triangle, C=180°AB<180°A\angle C=180\degree-\angle A-\angle B<180\degree-\angle A. We will consider two cases about A\angle A

Case AA, A90°\angle A \ge 90\degree

C<180°A90°\angle C<180\degree-\angle A\le 90\degree. Then, C\angle C can only be sin1(ABsinABC)\sin ^{ -1 }{ \left(\dfrac { AB\sin { \angle A } }{ BC }\right) }, so there are only 11 possible triangle similarly. Therefore, we can conclude that when A90°\angle A \ge 90\degree, SSASSA is true.

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

Case BB, A<90°\angle A < 90\degree

C<180°A\angle C<180\degree-\angle A. It is true when C=sin1(ABsinABC)\angle C=\sin ^{ -1 }{ \left(\dfrac { AB\sin { \angle A } }{ BC }\right) }. Because of we want to have 11 possibility of C\angle C, we need to ignore the other possibility of C\angle C, which we get the inequality below:

90°>180°sin1(ABsinABC)180°Asin90°>sin(180°sin1(ABsinABC))sin(180°A)ABsinABCsinAABBC90\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 ABBCAB\ge BC, SSASSA is true.

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

  • BCAB=sinA\dfrac { BC }{ AB }=\sin { \angle A }
  • A90°\angle A \ge 90\degree
  • ABBCAB\ge BC

I hope it will help you guys! Please comment below if you had something to say about this article. Thank you!

Note by Isaac Yiu Math Studio
2 weeks, 4 days ago

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