Question 1105541
<pre><b><font size = 5>
Since there is no accompanying diagram, I drew one.  It may not
be turned the same way as the one in your book.

{{{drawing(400,400,-3,5.5,-3,5.5,
line(0,0,3,0),locate(3.05,0.06,C),
circle(-2,-2.383507185,.05), locate(-2,-2.4,D),
locate(-.2,.27,B), locate(2.3,.45,"75°"),
locate(1.7,2.6,A),circle(4,4.76701437,.05),
locate(4,4.76701437,E),locate(2.2,2.4,"?°"),
locate(-.17,-.13,"130°"),
red(arc(0,0,1.3,-1.3,230,360),arc(3,0,1.7,-1.7,115,180),

arc(1.928362829,2.298133329,1.5,-1.5,300,410)

),
line(-2,-2.383507185,4,4.76701437),line(3,0,1.928362829,2.298133329) )}}}

&#8736;CBD and &#8736;ABC form a linear pair so m&#8736;ABC = 180°-130° = 50°

{{{drawing(400,400,-3,5.5,-3,5.5,
line(0,0,3,0),locate(3.05,0.06,C),
circle(-2,-2.383507185,.05), locate(-2,-2.4,D),
locate(-.2,.27,B), locate(2.3,.45,"75°"),
locate(1.7,2.6,A),circle(4,4.76701437,.05),
locate(4,4.76701437,E),locate(2.2,2.4,"?°"),
locate(-.17,-.13,"130°"),
red(arc(0,0,1.3,-1.3,230,360),arc(3,0,1.7,-1.7,115,180),

arc(1.928362829,2.298133329,1.5,-1.5,300,410) ),

green(arc(0,0,1.8,-1.8,0,50)),

locate(.3,.35,"50°"),
line(-2,-2.383507185,4,4.76701437),line(3,0,1.928362829,2.298133329) )}}}

&#8736;CAE is an exterior angle of &#916;ABC.
The measure of an exterior angle is the sum of the measures
of the two remote interior angles.  The two remote interior
angles are &#8736;ABC and &#8736;C .

So m&#8736;CAE = 50°+75° = 125°

Edwin</pre></b></font>