Question 1166670
*[illustration 1166670.jpg].
<pre>
AD ∥ CF  is given

AF ∥ CD  because they are both ⊥ DE

AFCD is a parallelogram because both pairs of opposite sides are ∥.

Use the Pythagorean theorem on right triangles △ADE and △ABE, and
we get DE = 144 an BE = 11.

△BEF ∽ △BDC  because  EF ∥ CD

{{{BE/BF=BD/BC}}} corresponding parts of similar triangles are in the
                  same ratio.

{{{BE/BF=(BE+DE)/(BF+CF)}}}

{{{11/BF=(11+144)/(BF+156)}}}

Solve that for BF and get

{{{BF=143/12}}}

Then using the Pythagorean theorem in rt. △BEF, we get

{{{EF=55/12}}}

Draw the altitude DG of of △BCD, DG ⊥ BC, where DG is 
the height of the trapezoid ABCD.

rt. △BDG ∽ rt. △BEF

{{{DG/BD=EF/BF}}}

{{{DG/(11+144)=(55/12)/(143/12)}}}

Solve that for DG and get

{{{DG=775/13}}}

{{{DG=59&8/13}}}

Edwin</pre>