now, we know that B and F are midpoints, namely BF is a midsegment of the triangle, and therefore BF || AE, so, the base AE is twice BF.
[tex]\bf 2BF=AE\implies 2(23)=5x-4\implies 46=5x-4\implies 50=5x \\\\\\ \cfrac{50}{5}=x\implies \boxed{10=x}[/tex]
