I want proof of SSS congruence rule in Geometry. I have tried to prove it by taking that all angles are equal i.e 60 degree. Then I applied SAS rule of congruence to prove it
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Why are all angles $60$ degree? – Apr 08 '15 at 06:23
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because sum of all angles in a triangle is 180 degree and since all angles in an equilateral triangle are equal, each must be 60 degree – A D Dechaudhari Apr 08 '15 at 09:39
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1@ADDechaudhari Not the sides of one triangle are supposed to be equal here. Instead, the corresponding sides of two triangles are equal. – Hagen von Eitzen Apr 08 '15 at 16:33
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You are right. I have solved the problem by doing some construction – A D Dechaudhari Apr 09 '15 at 07:48
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Two triangles are congruent if their corresponding sides are equal in length and their corresponding angles are equal in size.
If you are given that corresponding sides are equal in length, you can easily apply the Cosine Rule and obtain that each of the corresponding angles are also equal. Hence the two triangles are congruent.
For a more basic proof, refer here.
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1@zahbaz What I meant was the Law of Cosines, but the Law of Sines would probably work as well. – Arpan Apr 08 '15 at 06:31
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This is a problem for std IX where trigonometry is not introduced. so please prove it without cosine rule – A D Dechaudhari Apr 08 '15 at 09:43
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Hint: In how many points can two circles intersect?
Let $ABC,DEF$ be two triangles with $|AB|=|DE|\ne0$, $|BC|=|EF|$, $|AC|=|DF|$. Since a translation can transport $D\to A$, we may assume wlog. that $D=A$. Then a rotation around $A$ can take $E$ to the ray $AB$; since $|DE|=|AB|$, this rotation would take $E\to B$, so that we may assume wlog. that additionally $B=E$. Now $C,F$ are two points with the same distance to $A$, hence are on a circle around $A$. Similary, they are both on a circle around $B$: They are points of intersection of two distinct circles. If $F=C$, we are done. If $F\ne C$, then reflecting the figure at $AB$ leaves $A,B$ fixed as well as the circles having these points as center, hence permutes the points of intersection. There are only two such points of intersection and the permutation is non-trivial, hence it takes $F\to C$, thus showing the desired congruence.
(While the above argument may be quite persuading, the details depend on the axiom system actually used.)
Hagen von Eitzen
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Could you elaborate a little more and show how this is relevant to the question? – Arpan Apr 08 '15 at 09:50