Question 1163536
<br>
A sketch (NOT to scale)....<br>
AB=AC, so triangle ABC is isosceles;
BD=BE, so triangle BDE is isosceles;
AF=DF, so triangle AFD is isosceles.<br>
{{{drawing(600,600,0,100,0,100
,line(30,60,93.5,60),line(0,0,45,90),line(0,0,93.5,60),line(45,90,68,43.7)
,locate(45,93,A),locate(26,64,B),locate(62,64,C),locate(0,6,D),locate(95,64,E),locate(70,43,F)
)}}}<br>
Let x be the measure of angle ABC.<br>
Then x is the measure of angle ACB (isosceles triangle ABC).<br>
Then the measure of angle ECF is also x (vertical angle).<br>
Angle BAC is 180-2x (angle sum of triangle ABC).<br>
Angle ADF is 180-2x (isosceles triangle AFD).<br>
Angle BED is 180-2x (isosceles triangle DBE).<br>
Angle CFE is x (angle sum of triangle CEF).<br>
Angle AFD is 180-x (supplement to angle CFE).<br>
Angle DBC is 180-x (supplement to angle ABC).<br>
Now here is the sketch (still not to scale):<br>
{{{drawing(600,600,0,100,0,100
,line(30,60,93.5,60),line(0,0,45,90),line(0,0,93.5,60),line(45,90,68,43.7)
,locate(45,93,A),locate(26,64,B),locate(62,64,C),locate(0,6,D),locate(95,64,E),locate(70,43,F)
,locate(35,64,"x"),locate(54,64,"x"),locate(40,80,180-2x),locate(30,57,180-x),locate(54,57,180-x),locate(63,60,x),locate(80,60,180-2x),locate(68,48,x),locate(58,48,180-x),locate(6,12,180-2x)
)}}}<br>
Now look at quadrilateral DBCF.  The angle sum is<br>
FDB + DBC + BCF + CFD = (180-2x)+(180-x)+(180-x)+(180-x) = 720-5x<br>
But the angle sum of any quadrilateral is 360 degrees:<br>
{{{720-5x = 360}}}
{{{5x = 360}}}
{{{x = 72}}}<br>
ANSWER: The measure of angle ABC is 72 degrees.<br>
Note that, with a measure of 72 degrees for angle ABC, the problem description defines part of the figure of a regular 5-pointed star.<br>
Viewing the given not-to-scale sketch as part of a picture of a regular 5-pointed star makes it easy to see that all of the conditions of the problem as stated are satisfied.<br>
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Comment to tutor @MathTherapy....<br>
Nice, elegant solution, achieved by defining x to be the measure of angle BDE....<br>