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
5 months, 3 weeks ago

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The sum of the measures of the inside points of a triangle in Euclidean space is constantly 180 degrees. This reality is comparable to Euclid's parallel propose. This allows determination of the cheap assignment help proportion of the third angle of any triangle given the proportion of two angles. An exterior angle of a triangle is an angle that is a direct pair (and thus advantageous) to an interior angle.

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You might be tempted to think that given two sides and a non-included angle is enough to prove congruence. But there are two triangles possible that have the same values, so SSA is not sufficient to prove congruence. I am a content writer if you want uk assignment writing service contact me I am here to assist you.

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