Q:

Missing Delimiter (.inserted) error. How do I fix this?

\begin{proof}
We are given that there exists a quadrilateral $ABCD$ on a sphere. Let us create a line segment $AC$ that divides our quadrilateral into two triangles. Thus, $\angle{A}$ and $\angle{C}$ have been divided by our line segment $AC$. Let the portion of $\angle{A}$ and $\angle{C}$ on the same side of $AC$ as $B$ be called $\angle{A_1}$ and $\angle{C_1}$. Also, Let the portion of $\angle{A}$ and $\angle{C}$ on the same side of $AC$ as $D$ be called $\angle{A_2}$ and $\angle{C_2}$. Therefore, 

\begin{align*}
\left|ABCD|\right &=\left|\Delta{ABC}|+|\Delta{ACD}|\right\\
\ &= \left p^2(\angle{A_1}+\angle{B}+\angle{C_1}-\pi)+p^2(\angle{A_2}+\angle{C_2}+\angle{D}-\pi)\right\\
\ &= \left p^2(\angle{A_1}+\angle{B}+\angle{C_1}+\angle{A_2}+\angle{C_2}+\angle{D}-2\pi)\right.
\end{align*}

However, \[\angle{A_1}+\angle{A_2}=\angle{A}\] and \[\angle{C_1}+\angle{C_2}=\angle{C}\] \[\left|ABCD|\right=\left p^2(\angle{A}+\angle{B}+\angle{C}+\angle{D}-2\pi)}\right\].
To generalize furthur the area of any $n$-sided polygon on the sphere is as follows, \[\left{Area of an N-sided Polygon}\right=\left(sum of the angles-(n-2)\pi\right\]

\end{proof}

This is code that is producing the missing delimiter error, it says the problem is occurring somewhere with the align environment. Anyone have suggestions?

A:

the problem here is how the \left and \right commands are used.
each of these commands always requires either an actual delimiter --
parenthesis, bracket, etc. -- or a "placeholder", namely a period.
in the first line of the alignment, the \right is placed after the |;
it should be before.
so at the beginning of the second and third lines (and perhaps elsewhere, i didn't check), \left is not followed by any delimiter, it's not clear what
delimiter is to be "matched".  perhaps the opening parenthesis after \p^2?
here is the reformulated display:
\begin{align*}
\left|ABCD\right| &=\left|\Delta{ABC}\right|+\left|\Delta{ACD}\right|\\
\ &= p^2\left(\angle{A_1}+\angle{B}+\angle{C_1}-\pi)+p^2(\angle{A_2}
   +\angle{C_2}+\angle{D}-\pi\right)\\
\ &= p^2\left(\angle{A_1}+\angle{B}+\angle{C_1}+\angle{A_2}
   +\angle{C_2}+\angle{D}-2\pi\right).
\end{align*}

actually, since most of the symbols between the delimiters aren't taller
than the normal text, \left and \right aren't really needed.
in the paragraph following the align*, i think you're confusing \[
and \] with the actual square brackets.
the "escaped" forms indicate the beginning and end of a one-line math
display.  no backslashes should be used if an actual bracket is intended.
however, a typeset brace must be entered preceded by a backslash.  and
within a math environment, normal text (the "Area of ...") needs to be so
indicated, by \text{...}, and within that string, any small math
expressions need to be returned to math mode by $...$.
here is that paragraph reformulated:
However, [\angle{A_1}+\angle{A_2}=\angle{A}] and
[\angle{C_1}+\angle{C_2}=\angle{C}]
[\left|ABCD\right|= p^2\left(\angle{A}+\angle{B}+\angle{C}
 +\angle{D}-2\pi\right)}].
To generalize further the area of any $n$-sided polygon on the sphere is as follows,
\[\left\{\text{Area of an $N$-sided Polygon}\right\}
 =\left\{\text{sum of the angles $-(n-2)\pi$}\right\}\]

i may have misunderstood your intent in the paragraph following the display;
if you really did intend these bracketed expressions to be displayed, then
instead of coding each one as a separate display, you should use the
gather* environment.
you don't say what document class or theorem package you are using.  however,
in any event, you shouldn't ever leave a blank line before \end{proof} --
that will guarantee that the "tombstone" is always set on a line by itself.
also, if you are using amsthm, you can insert \qedhere before
the closing \] to place the "tombstone" on the last line of the display.

