Triangles

By the angle bisector theorem, $\dfrac{AC}{CB}=\dfrac{AD}{DB}=\dfrac{3}{4}$ . Let $AC=3x$ and $CB=4x$ . Then by pythagorean theorem, we have,

$(3x)^2+(4x)^2=7^2$ $\implies$ $9x^2+16x^2=49$ $\implies$ $x=1.4$

It follows that, $AC=3(1.4) = 4.2$ and $CB=4(1.4) = 5.6$ .

By the angle bisector theorem, $\dfrac{AC}{AB}=\dfrac{CE}{EB}$ , substituting we get

$\dfrac{4.2}{7}=\dfrac{CE}{5.6-CE}$ $\implies$ $CE=2.1$

By pythagorean theorem,

$AE=\sqrt{4.2^2+2.1^2}=$ $\color{#D61F06}\boxed{4.6957}$