## LaTeX forum ⇒ Math & Science ⇒ Beginner : i dont know the errors in my Latex document Topic is solved

Information and discussion about LaTeX's math and science related features (e.g. formulas, graphs).
hariekd
Posts: 2
Joined: Sat Nov 28, 2015 12:55 pm

### Beginner : i dont know the errors in my Latex document  Topic is solved

I am a new latex user. I have prepared a maths proof for my students. Even i get the out put, the source file shows so many errors. pls indicate me about the errors.
\documentclass{article}\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}

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Stefan Kottwitz
Posts: 9604
Joined: Mon Mar 10, 2008 9:44 pm
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

$\therefore \triangle APQ \sim \triangle ABC$

• 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: \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. \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