Large link border with hyperref

The linkborder adapts to the surrounding box. As you have a large cases in the current line, you get a large link box. You can avoid it by putting an additional \mbox around the ref:

\documentclass[12pt,a4paper]{article}
\usepackage{hyperref}
    \hypersetup{linkcolor = {0 .8 1}}
\begin{document}

\section{xxx}\label{test}

\rule{2mm}{1cm} high: \ref{test}, short: \mbox{\ref{test}}.
\end{document}

enter image description here


I think it's not a good idea to use inline math mode for the entire proof, both because super-tall lines are ugly and because the use of inline math doesn't do a good job of visually highlighting what the main logical step is.

I'm afraid I'm not fluent in Portuguese, but assuming I understood the material correctly, I think it's better to use \angle than \hat to denote, well, angles.

For the following screenshot, the code was compiled under pdfLaTeX. Results with either LuaLaTeX or XeLaTeX are virtually the same.

enter image description here

\documentclass[12pt,a4paper]{article}
\usepackage[hmargin=2cm]{geometry}
\usepackage{amsmath, amsthm, amssymb}
\usepackage[portuguese]{babel} % or 'brasilian'?

\usepackage{ifluatex,ifxetex}
\ifluatex
   \usepackage{unicode-math,lualatex-math}
\else\ifxetex
   \usepackage{unicode-math}
\else
   \usepackage[utf8]{inputenc}
   \usepackage[T1]{fontenc}
\fi\fi

\usepackage{hyperref}
\hypersetup{linkcolor = {0 .8 1}}
\usepackage[nameinlink]{cleveref}
\theoremstyle{definition}
\newtheorem{ax}{Axioma}

\begin{document}
\begin{ax}\label{ax:1}
Dados dois triângulos $ABC$ e $EFG$, se $AB\equiv EF$, $AC\equiv EG$ 
e $\angle{A}\equiv\angle{E}$, então $ABC\equiv EFG$.
\end{ax}    

\noindent\dots

\begin{proof}
Seja $ABC$ e $EFG$ dois triângulos de correspondência $ABC\leftrightarrow EFG$. 
\dots\ Assim, por hipótese tem-se $AB\equiv EF$ e por construção tem-se 
$AD\equiv EF$. 
\[
\text{Pelo \cref{ax:1}: }
\begin{cases}
    AC\equiv EF \\
    \angle E\equiv C\angle AD \\
    AD\equiv EF
\end{cases} 
\quad \Rightarrow \triangle ACD \equiv \triangle EFG.
\qedhere % take care to place the QED symbol
\]
\end{proof}
\end{document}