Check the picture below, on the left side.
since the triangle ABC is an isosceles with twin sides, if we drop a line bisecting the angle at the vertex A, we end up with a perpendicular line that cuts the "base" in two equal halves, so then
[tex]\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=\stackrel{hypotenuse}{50}\\ a=\stackrel{adjacent}{14}\\ b=opposite\\ \end{cases} \\\\\\ \sqrt{50^2 - 14^2}\implies \sqrt{2500 - 196}=b\implies \sqrt{2304}=b\implies 48=b[/tex]
as you can see in the picture in red, now let's find CD.
[tex]\stackrel{\textit{pythagorean theorem}}{\sqrt{52^2 - 48^2}=14}+CD\implies \sqrt{2704-2304}=14+CD \\\\\\ \sqrt{400}=14+CD \implies 20=14+CD\implies 6=CD[/tex]
