[Tex/LaTex] How to shift an array further left

Question

I want to move an array about 3 or 4 cm further to the left. Knowing how to align it on the left side of the page could be useful too.

The code is:

\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
    \section*{Question 10}
    \begin{figure}[h!]
        \includegraphics[width=\linewidth]{"D:/.../question 10".png}
    \end{figure}
    Let $K\hat{P}L = x$ and $K\hat{Q}M = y$. Required to prove: $\triangle KPL ||| \triangle KMQ$.
    \linebreak
    \[ \begin{array}{rll}
        & O\hat{M}Q=y & (\text{isoc.} \triangle, OQ=OM)\\
        \Rightarrow \ & P\hat{O}M = 2y & (\text{ext.} \angle \triangle)\\
        \Rightarrow \ & \text{reflex} \angle P\hat{O}M =360^{\circ}-2y & (\angle \text{'s around a point})\\
        \Rightarrow \ & 2P\hat{L}M = 360^{\circ}-2y & (\angle \text{at cntr}= \times 2)\\
        \Rightarrow \ & P\hat{L}M = 180^{\circ}-y\\
        \Rightarrow \ & K\hat{L}P = y & (\angle \text{'s on st. line})\\
    \end{array} \]
    So, in $\triangle KPL$ and $\triangle KMQ$
    \[ \begin{array}{rll} 
        1. & \hat{K} \text{ is common}\\
        2. & K\hat{L}P=y=K\hat{Q}M\\
        \vspace{4mm}
        \therefore & \triangle KPL ||| \triangle KMQ & (A.A.)
    \end{array} \]
\end{document}

I want to align the second array further left.

The document looks like this:

Secondly, it seems as if the \vspace{2mm} command is not working. How can I get a bit more space in-between the last two lines?

Thirdly, why is the word "Let" and the word "So" not aligned?

Answer 1

You can use the missile flalign* to left align the equations.

\begin{flalign*}
& equation &
\end{flalign*}

You can align the parts using the & method of flalign* itself, but I left it as an exercise.

You can add the vertical space using \\[4mm]

Let is indented so not aligned with So. Use \noindent.

Code:

\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
    \section*{Question 10}
    \begin{figure}[h!]
        \includegraphics[width=0.2\linewidth]{example-image}
    \end{figure}
    \noindent Let $K\hat{P}L = x$ and $K\hat{Q}M = y$. Required to prove: $\triangle KPL ||| \triangle KMQ$.
    \begin{flalign*}
    &\begin{array}{rll}
        & O\hat{M}Q=y & (\text{isoc.} \triangle, OQ=OM)\\
        \Rightarrow \ & P\hat{O}M = 2y & (\text{ext.} \angle \triangle)\\
        \Rightarrow \ & \text{reflex} \angle P\hat{O}M =360^{\circ}-2y & (\angle \text{'s around a point})\\
        \Rightarrow \ & 2P\hat{L}M = 360^{\circ}-2y & (\angle \text{at cntr}= \times 2)\\
        \Rightarrow \ & P\hat{L}M = 180^{\circ}-y\\
        \Rightarrow \ & K\hat{L}P = y & (\angle \text{'s on st. line})\\
    \end{array} &
    \end{flalign*}
    So, in $\triangle KPL$ and $\triangle KMQ$
    \begin{flalign*}
     &\begin{array}{rll}
        1. & \hat{K} \text{ is common}\\
        2. & K\hat{L}P=y=K\hat{Q}M\\[4mm]
        \therefore & \triangle KPL ||| \triangle KMQ & (A.A.)
    \end{array} &
    \end{flalign*}
\end{document}