to one another, and likewise their sides CB, DB, which are terminated at B equal to one another.
Join CD. In the case in which the vertex of each triangle is without the other triangle;
because AC is equal to AD,[Hypothesis.
the angle ACD is equal to the angle ADC.[I. 5.
But the angle ACD is greater than the angle BCD,[Ax. 9.
therefore the angle ADC is also greater than the angle BCD;
much more then is the angle BDC greater than the angle BCD.
Again, because BC is equal to BD,[Hypothesis.
the angle BDC is equal to the angle BCD.[I. 5.
But it has been shewn to be greater; which is impossible.
But if one of the vertices as D, be within the other triangle ACB, produce AC, AD to E, F.
Then because AC is equal to AD, in the triangle ACD, [Hyp.
the angles ECD, FDC, on the other side of the base CD, are equal to one another. [I. 5.
But the angle ECD is greater than the angle BCD, [Axiom 9.
therefore the angle FDC is also greater than the angle BCD;
much more then is the angle BDC greater than the angle BCD.
Again, because BC is equal to BD, [Hypothesis.
the angle BDC is equal to the angle BCD. [I. 5.
But it has been shewn to be greater; which is impossible.
The case in which the vertex of one triangle is on a side of the other needs no demonstration.
Wherefore, on the same base &c. q.e.d.
PROPOSITION 8. THEOREM.
If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their
