I can find the proof(involving equations) of all congruence criterias of triangles except the S.A.S. rule . Everywhere, the proof of S.A.S. is given as they place one triangle on another triangle and find that all things coincides , which I can't digest logically. Isn't there any proof of this which involves equations?
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$\begingroup$ You can't prove "all" congruence criteria. One of them has to be taken as an assumed axiom to get things started. Traditionally, SAS takes this honor. (It's an interesting exercise in Foundational Geometry to start w/any of SAS, SSS, ASA and prove the other two.) I think of the intuition like this: If you and I take matching pairs of sticks and incline them at matching angles, then it "should" be true (and hence axiomatic) that if we each join the ends of our sticks we'll get matching triangles. (See also comments after this answer of mine.) $\endgroup$– BlueMar 5, 2022 at 12:19
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$\begingroup$ I saw . What I learnt is that accept any one of criteria without proof . $\endgroup$– Get_ MathsMar 5, 2022 at 14:13
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$\begingroup$ See my answer: math.stackexchange.com/q/4166887 $\endgroup$– gpassanteMar 6, 2022 at 15:19
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$\begingroup$ I saw , it's very high level stuff . $\endgroup$– Get_ MathsMar 6, 2022 at 15:38
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$\begingroup$ My answer is a bit long, but it is basically just based on the study of the isometry $f$. If I may make a suggestion, you can try to transport that study to an analogous function for your problem. $\endgroup$– gpassanteMar 6, 2022 at 15:56
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1 Answer
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If we know 2 sides and the angle between them, just use the law of cosine to find the third side
LAW OF COSINE
$a^2=b^2+c^2-2bc \cos A$
$b$ and $c$ are the know sides and angle $A$ is the angle between them.
And then it can be converted into SSS congruence rule
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$\begingroup$ But first you should proof SSS . You cannot use SSS to prove SAS because SSS itself is proved using SAS . It's just circular. $\endgroup$ Mar 5, 2022 at 10:05
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$\begingroup$ The law of cosine is $a^2=b^2+c^2-2bc\cos A.$ $\endgroup$ Mar 5, 2022 at 11:04
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$\begingroup$ You can't use the theorums which are based on this theorum to prove this . Prove SAS without using other congruence criterias. $\endgroup$ Mar 5, 2022 at 14:11