2

I have tried searching in many places for some good proofs of these theorems but couldn't find them anywhere . Even my math teacher cannot explain it to me and says that these theorems just work.

I am a ninth grader so please try to explain in simple terms .

MayankJain
  • 521
  • Have you explored Google : https://www.google.co.in/?gws_rd=cr&ei=2NY1UsOcD8L_rQexnoDYAg#q=triangle%20congruence%20theorems%20proof – lab bhattacharjee Sep 15 '13 at 18:27
  • I am not sure what you mean by RHS – Mark Bennet Sep 15 '13 at 18:35
  • @labbhattacharjee I did try finding the proof with the exact search terms in your link but couldn't find a proof that satisfied me . – MayankJain Sep 15 '13 at 18:48
  • @MarkBennet RHS - If the hypotenuse and one side of a right angled triangle are equal respectively to the corresponding hypotenuse and one side of another right angled triangle, then the two right angled triangles are congruent. – MayankJain Sep 15 '13 at 18:48
  • Try this version of Euclid http://aleph0.clarku.edu/~djoyce/java/elements/elements.html - you want Book I propositions eg $4$ and $26$ – Mark Bennet Sep 15 '13 at 19:34

1 Answers1

0

With "Angle-Side-Angle" we know the third angle and use the sine rule $\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}(=2R)$ to determine the unknown sides.

With "Side-Angle-Side" we use the cosine rule to determine the third side opposite the given angle $a^2=b^2+c^2-2bc \cos A$, and then the other angles from the same formula.

With "Side-Side-Side" we use the cosine rule to determine the angle opposite each side.

This always puzzled me until I realised that the sine and cosine rules effectively encoded the uniqueness of these constructions. No-one ever pointed out the connection, which is obvious when you see it.

In a right-angled triangle if we know the hypotenuse and one side, we find the other side using Pythagoras (a stripped down cosine rule), and then we have three sides and an angle and have reduced to one of the other cases.

In the general case SSA the cosine rule gives a quadratic for the third side, and there are two solutions. With a right-angled triangle the two solutions coincide.

Mark Bennet
  • 100,194
  • I am sorry but we are not taught trigonometry in 9th grade here. I do know some basic terms such as sin , cos , tan etc . but I am not familiar with these rules that you are talking about . I'll try to read about them but can you explain it in some simple terms ? – MayankJain Sep 15 '13 at 18:53
  • @Mayank The formula for the area of a triangle is $\frac 12 fh$ where $f$ is the base and $h$ is the height. It is standard to name the vertices of a triangle $A, B, C$ and to use the same letters for the angles, while the sides opposite are respectively $a, b, c$ so the vertex $A$ is where sides $b$ and $c$ meet. Take $c$ as the base of the triangle - then the height is $b\sin A$ (this is a basic use of the sine function. - so the area is $\frac 12 bc \sin A$. Calculating at $B$ we get the same area is $\frac 12 ac \sin B$ and putting the two equal we get $\frac a{\sin A}=\frac b{sin B}$ – Mark Bennet Sep 15 '13 at 19:08
  • I still didn't get it . ok so we know that $\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}(=2R)$ but how can we determine the unknown sides using this ? What is R ? I think this is too advanced for me :( – MayankJain Sep 15 '13 at 19:21
  • @Mayank The cosine formula is a generalisation of Pythagoras, and less easy to explain in a comment - I suggest looking it up eg http://mathproofs.blogspot.co.uk/2006/06/law-of-cosines.html. Note the original ideas go back to Euclid - his proposition I.4 is SAS and proposition I.26 is ASA - and these are done without trigonometry (I have Heath's translation). Euclid is worth a look if you are interested, to see how the results build up to a conclusion. – Mark Bennet Sep 15 '13 at 19:21
  • @Mayank Ignore $R$ - it is the circumradius of the triangle. If I know ASA then I know all the angles including the angle $A$ and the side $a$. I then calculate $b=\frac {a\sin B}{\sin A}$ and $c=\frac {a\sin C}{\sin A}$ – Mark Bennet Sep 15 '13 at 19:25
  • The fact that $ASA$ is Euclid I.26 suggests a fair amount of work is required to establish it without using trigonometry - this is because some care has to be taken over the definition of angles. Of course the trigonometric functions encode information about similar triangles, and assume that angles behave as we expect. – Mark Bennet Sep 15 '13 at 19:28
  • I'm fairly certain that proofs of the law of sines (or cosines) depend, at some step, on theorems which require ASA for their proofs. Therefore, this proof is very probably circular reasoning. – Mark Fischler Jul 08 '16 at 21:29