Hi,
welcome to the forum!
Very good that you started with LaTeX. I recommend to read an introductory text. For example, my book
LaTeX Beginner's Guide, or a free text such as
LaTeX for Complete Novices.
By the way, this weekend my publisher sells my two LaTeX ebooks, so also the
LaTeX Cookbook, (all ebooks) with 50% discount (
link)
- There are problems with inline math mode in the text, that is, math formulas within normal text. Start with a $, later end with a $. A rule of thumb, symbols are in math mode too. So, for example write
Code: Select all
$\therefore \triangle APQ \sim \triangle ABC$
- Don't use $ within
align, because this is already (displayed) math mode.
- Load the amsmath package for extended math support.
Here is the corrected error-free code, but some more things can be improved:
Code: Select all
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\centerline{\sc \Large Standard IX - unit 9 Similar Triangles}
\vspace{.5pc}
\centerline{\it (Second Question from Page Number 133)}
\vspace{2pc}
\textbf{In a trianlge, a line is drawn parallel to one side and a small triangle is cut off. If the Parallel line drawn is through the mid-point of one side, then how much of the area of the original triangle is the area of the small triangle?}\\
Proof:- Consider $\triangle APQ$ and $\triangle ABC$
\begin{enumerate}
\item $\angle APQ =$ and $\angle ABC$ (Corresponding Angles)
\item $\angle AQP =$ and $\angle ACB$ (Corresponding Angles)
\end{enumerate}
As the two angles of $\triangle APQ$ are equal to the two angles of $\triangle ABC$, these two triangles will be similar.\\
$\therefore \triangle APQ \sim \triangle ABC$\\
As $P$ and $Q$ are the midpoints of $AB$ and $AC$,\\
$AP=\frac{1}{2}AB$ and $AQ=\frac{1}{2}AC$\\
$\therefore \frac{AQ}{AC}=$ $\frac{AP}{AB}=$ $\frac{PQ}{BC}=$ $\frac{1}{2}$
\begin{equation}
ie, AP= \frac{1}{2}AB
\end{equation}
Consider $\triangle APR$ and $\triangle ABD$
\begin{enumerate}
\item $\angle APR =$ and $\angle ABD$ (Corresponding Angles)
\item $\angle ARP =$ and $\angle ADB$ (Corresponding Angles)
\end{enumerate}
As the two angles of $\triangle APR$ are equal to the two angles of $\triangle ABD$, these two trianlges will be similar.\\
$\therefore \triangle APR \sim \triangle ABD$
\newline
$\therefore \frac{AR}{AD}$ = $\frac{AP}{AB}$= $\frac{1}{2}$ (We have proved $\frac{AP}{AB}$= $\frac{1}{2}$)\\
\begin{equation}
ie, AR=\frac{1}{2}AD
\end{equation}
\begin{align}
Area of \triangle APQ
& = \frac{1}{2} X PQ X AR \\
& = \frac{1}{2} X \frac{1}{2}BC X \frac{1}{2}AD (By using Equation 1 and 2) \\
& = \frac{1}{4} X \frac{1}{2} X BC X AD \\
& = \frac{1}{4} X Area of \triangle ABC \\
\end{align}
\end{document}
For example, don't end a text line by
\\. This is a command just for ending lines in a table or a multi-line math formula. An empty line is sufficient as a paragraph break.
It seems that you (mis)use
\\ to get a space between paragraph. For this purpose, you could load the
parskip package and remove the
\\ in normal text.
Code: Select all
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{parskip}
\begin{document}
\centerline{\sc \Large Standard IX - unit 9 Similar Triangles}
\vspace{.5pc}
\centerline{\it (Second Question from Page Number 133)}
\vspace{2pc}
\textbf{In a triangle, a line is drawn parallel to one side and a small triangle is cut off. If the Parallel line drawn is through the mid-point of one side, then how much of the area of the original triangle is the area of the small triangle?}
Proof:- Consider $\triangle APQ$ and $\triangle ABC$
\begin{enumerate}
\item $\angle APQ =$ and $\angle ABC$ (Corresponding Angles)
\item $\angle AQP =$ and $\angle ACB$ (Corresponding Angles)
\end{enumerate}
As the two angles of $\triangle APQ$ are equal to the two angles of $\triangle ABC$, these two triangles will be similar.
$\therefore \triangle APQ \sim \triangle ABC$
As $P$ and $Q$ are the midpoints of $AB$ and $AC$,
$AP=\frac{1}{2}AB$ and $AQ=\frac{1}{2}AC$
$\therefore \frac{AQ}{AC}=$ $\frac{AP}{AB}=$ $\frac{PQ}{BC}=$ $\frac{1}{2}$
\begin{equation}
ie, AP= \frac{1}{2}AB
\end{equation}
Consider $\triangle APR$ and $\triangle ABD$
\begin{enumerate}
\item $\angle APR =$ and $\angle ABD$ (Corresponding Angles)
\item $\angle ARP =$ and $\angle ADB$ (Corresponding Angles)
\end{enumerate}
As the two angles of $\triangle APR$ are equal to the two angles of $\triangle ABD$, these two trianlges will be similar.
$\therefore \triangle APR \sim \triangle ABD$
\newline
$\therefore \frac{AR}{AD}$ = $\frac{AP}{AB}$= $\frac{1}{2}$ (We have proved $\frac{AP}{AB}$= $\frac{1}{2}$)
\begin{equation}
ie, AR=\frac{1}{2}AD
\end{equation}
\begin{align}
Area of \triangle APQ
& = \frac{1}{2} X PQ X AR \\
& = \frac{1}{2} X \frac{1}{2}BC X \frac{1}{2}AD (By using Equation 1 and 2) \\
& = \frac{1}{4} X \frac{1}{2} X BC X AD \\
& = \frac{1}{4} X Area of \triangle ABC \\
\end{align}
\end{document}
Maybe you learned LaTeX from some old examples, which are not perfect. Reading a book or an introduction can help to get a better start, such as I meant above.
Stefan
