Page:First six books of the elements of Euclid 1847 Byrne.djvu/41

N the ſame baſe (), and on the ſame ſide of it there cannot be two triangles having their conterminous ſides ( and , and ) at both extremities of the baſe, equal to each other.

When two triangles ſtand on the same baſe, and on the ſame ſide of it, the vertex of the one ſhall either fall outſide of the other triangle, or within it; or, laſtly, on one of its ſides.

If it be poſſible let the two triangles be conſtructed ſo that $\scriptstyle {\left\{{\begin{matrix}\ \end{matrix}}\right.}$=
=$\scriptstyle {\left.{\begin{matrix}\ \end{matrix}}\right\}\,}$, then

	draw and,
	$=$ (pr. 5.)
	$\therefore$ $\sqsupset$ and
	$\therefore$ $\sqsupset$	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \ \end{matrix}}\right\}\,}$	which is abſurd,
but (pr. 5.) $=$

therefore the two triangles cannot have their conterminous ſides equal at both extremities of the baſe.

Q.E.D.