Question 1105661
<pre>{{{drawing(400,4800/19,-2,17,-2,10,

locate(0,0,B), locate(15,0,C), locate(0,8.8,A),locate(15,8.8,D),
locate(960/289,512/289,E), locate(3375/289-.3,1800/289+.7,F), 
line(0,0,15,0),
line(15,8,15,0),
line(0,8,15,8),
line(0,0,0,8),
line(0,0,15,8),
line(0,8,960/289,512/289),
locate(7.4,0,15), locate(15.1,4,8),
line(3375/289,1800/289,15,0) )}}}

By applying the Pythagorean theorem to right &#916;BCD,
BD = 17

&#916;BCD &#8765; &#916;CFD because they are right triangles with a common
acute angle, so

{{{FD/DC}}}{{{""=""}}}{{{ DC/BD}}}
{{{FD/8}}}{{{""=""}}}{{{ 8/17}}}
{{{17*FD}}}{{{""=""}}}{{{ 8*8}}}
{{{17*FD}}}{{{""=""}}}{{{ 64}}}
{{{FD}}}{{{""=""}}}{{{ 64/17}}}

&#916;CFD &#8773; &#916;AEB is easy to show.

So FD = EB  by cpct

{{{FD}}}{{{""=""}}}{{{EB}}}{{{""=""}}}{{{ 64/17}}}

BD = EB + EF + FD
{{{17}}}{{{""=""}}}{{{ 64/17+EF+64/17}}}
{{{17}}}{{{""=""}}}{{{ 128/17+EF}}}

Multiply though by 17

{{{289}}}{{{""=""}}}{{{ 128+17*EF}}}

{{{161}}}{{{""=""}}}{{{ 17*EF}}}

{{{161/17}}}{{{""=""}}}{{{ EF}}}

Edwin</pre>