Why can't there be an axiom of congruency of triangles as A.S.S. similar to R.H.S.?

I think that, if we consider 2 triangles, whose one angle, the side opposite to that angle and any other side are equal, then the two triangles are congruent. If I am wrong, can someone please construct 2 non-congruent triangles with these conditions?

1 Answer
Jul 18, 2016

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(details below)

Explanation:

If #C# is the center of a circle, the #abs(CB)=abs(CD)#

By construction
#color(white)("XXX")/_BAC=/_DAC#

In triangles #triangle BAC# and #triangle DAC#
#color(white)("XXX")/_BAC=/_DAC#
#color(white)("XXX")abs(AC)=abs(AC)#
and
#color(white)("XXX")abs(CB)=abs(CD)#

So we have an A.S.S. arrangement
but
#color(white)("XXX")triangle ACB# is not congruent to #triangle ACD#