Question 1189508
<pre>
{{{drawing(400,12800/63,-3,28.5,-3,13,

triangle(0,0,3.907311285,9.205048535,25.59304665,0),
triangle(0,0,3.907311285,9.205048535,10,0),
triangle(2,12.086682,10,0,10,0),triangle(2*1.953655642,8.660254038,2*1.953655642,12,2*1.953655642,8.660254038),locate(19.4,2,23^o),
locate(0,0,F), locate(10,0,A), locate(25.6,0,B), locate(4.4,10.2,C),locate(4.4,13,E) , locate(1.7,12,D),
red(arc(3.907311285,9.205048535,4,-4,302,336), arc(3.907311285,9.205048535,4,-4,90,124)) 


 )}}}

Since ∠BCF is a right angle, ΔBCF is a right triangle and
m∠FCG = 90<sup>o</sup> - 23<sup>o</sup> = 67<sup>o</sup>.

The fact that CE ⊥ AB, means that an extension of CE down to an 
extension of AB, the extensions will intersect at a right angle. So
we extend EC down to FA, which is a left extension of AB. Let G be 
the point where the extension of EC intersepts FA.

{{{drawing(400,12800/63,-3,28.5,-3,13,

triangle(0,0,3.907311285,9.205048535,25.59304665,0),
triangle(0,0,3.907311285,9.205048535,10,0),
triangle(2,12.086682,10,0,10,0),triangle(2*1.953655642,8.660254038,2*1.953655642,12,2*1.953655642,8.660254038),locate(19.4,2,23^o),
locate(0,0,F), locate(10,0,A), locate(25.6,0,B), locate(3.9,0,G),locate(4.4,10.2,C),locate(4.4,13,E) , locate(1.7,12,D),
green(line(3.907311285,0,3.907311285,9.205048535)),
red(arc(3.907311285,9.205048535,4,-4,302,336), arc(3.907311285,9.205048535,4,-4,90,124)), locate(1,2,67^o) 


 )}}}

Now that we extended EC to G we see that ∠DCE ≅ ∠ACG because
they are vertical angles. We also see that 

Since ∠CGF is a right angle, ΔCGF is a right triangle and
m∠GCF = 90<sup>o</sup> - 67<sup>o</sup> = 23<sup>o</sup>.

Let the common measure of ∠DCE, ∠ACG, and ∠ACB be x<sup>o</sup>

{{{drawing(400,12800/63,-3,28.5,-3,13,

triangle(0,0,3.907311285,9.205048535,25.59304665,0),
triangle(0,0,3.907311285,9.205048535,10,0),
triangle(2,12.086682,10,0,10,0),triangle(2*1.953655642,8.660254038,2*1.953655642,12,2*1.953655642,8.660254038),locate(19.4,2,23^o),
locate(0,0,F), locate(10,0,A), locate(25.6,0,B), locate(3.9,0,G),locate(4.4,10.2,C),locate(4.4,13,E) , locate(1.7,12,D),
green(line(3.907311285,0,3.907311285,9.205048535)),
red(arc(3.907311285,9.205048535,4,-4,302,336), arc(3.907311285,9.205048535,4,-4,90,124),
arc(3.907311285,9.205048535,4.5,-4.5,270,304)

), locate(1,2,67^o), locate(2.1,6,23^o),locate(4.3,7,x^o),locate(6.1,8,x^o) 


 )}}}

Since Angle BCF is a right angle,

{{{23^o+x^o+x^o}}}{{{""=""}}}{{{90^o}}}

{{{23^o+2x^o}}}{{{""=""}}}{{{90^o}}}

{{{2x^o}}}{{{""=""}}}{{{67^o}}}

{{{x^o}}}{{{""=""}}}{{{33.5^o}}}

Since the three angles of ΔBAC must have sum 180<sup>o</sup>,

m∠BAD + m∠ABC + m∠BCA = 180<sup>o</sup>

m∠BAD + 23<sup>o</sup> + 33.5<sup>o</sup> = 180<sup>o</sup>

m∠BAD + 56.5<sup>o</sup> = 180<sup>o</sup>

m∠BAD = 123.5<sup>o</sup>

Edwin</pre>