Disproving $A-S-S(Angle-Side-Side)$ congruence condition...

Question

I know that $A-S-S$$(Angle-Side-Side)$ congruence does not exist.But I cannot disprove it. Every time I draw a figure,I get two congruent triangles.

My Attempt-

So,we draw two lines such that $AB=DE$.Now we draw angles $\angle CAB=\angle FDB.$Now,By compass we take measurements such that $CB=FE.$By Cutting arcs on lines $A$ and $B$ we see that $AC=DF$ always.So,Now the two triangles are congruent by $SAS.$

Someone please help me out of this,with a diagram to disprove it.

Thanks a lot in advance.

Answer 1

Here is a counterexample that should be what you are looking for:

Answer 2

Look at your method but imagine that last side being done by setting the point of compasses, set to the given length, at E and striking an arc. There are three possible results. In might happen that the arc is simply NOT long enough to reach side AC. In that case, there is NO such triangle, It might happen that the arc cuts side AC in two different places. In that case there are two such triangles (this is the case that aras is giving). It might happen that the arc strikes side AC in exactly one place. In that case, side AC is tangent to the arc so the triangle is a right triangle.

Answer 3

Did you see that your example is a right triangle? Look at your method but imagine that last side being done by setting the point of compasses, set to the given length, at E and striking an arc. There are three possible results. In might happen that the arc is simply NOT long enough to reach side AC. In that case, there is NO such triangle, It might happen that the arc cuts side AC in two different places. In that case there are two such triangles (this is the case that aras is giving). It might happen that the arc strikes side AC in exactly one place. In that case, side AC is tangent to the arc so the triangle is a right triangle.