Check the picture below.
so, whenever we have a line coming from a vertex in a right triangle, like in this case, and drops to a perpendicular line, we end up with 3 similar triangles, as you can see in the picture, thus we can use proportions to get the sides, we'll use those proportions in the picture in blue.
[tex]\bf \stackrel{large}{\cfrac{14}{16}}=\stackrel{medium}{\cfrac{c}{14}}\implies \cfrac{7}{8}=\cfrac{c}{14}\implies 98=8c\implies \cfrac{98}{8}=c\implies \boxed{\cfrac{49}{4}=c} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{we know that}}{c+d=16}\implies d = 16 -c\implies d = 16-\cfrac{49}{4}\implies \boxed{d = \cfrac{15}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{small}{medium}\qquad \qquad \cfrac{e}{c}=\cfrac{d}{e}\implies e^2=cd\implies e=\sqrt{cd}\implies e = \sqrt{\cfrac{49}{4}\cdot \cfrac{15}{4}}[/tex]
[tex]\bf e = \sqrt{\cfrac{660}{16}}\implies e = \sqrt{\cfrac{4\cdot 165}{16}}\implies e = \sqrt{\cfrac{165}{4}} \\\\\\ e = \cfrac{\sqrt{165}}{\sqrt{4}}\implies \boxed{e = \cfrac{\sqrt{165}}{2}}[/tex]
