### MATH 113: Interior Angles of a Quadrilateral

In class we saw a proof that the sum of the measures of the exterior angles of a triangle is 360 degrees, and we also saw an illustration (not a proof) that this is true for all convex polygons.
Neatly write up and turn in a convincing argument that the sum of
the measures of the interior angles of *any* convex
quadrilateral is 360 degrees. Your argument should be between 1/2 and
2 pages long, including figures. You may assume that the sum of the
measures of the exterior angles of any convex quadrilateral is 360
degrees. Arguments based on studying sample quadrilaterals will
receive partial, but not full, credit.

You can find answers to this question on the internet and in geometry textbooks. If you do so, please cite the reference you used and be sure to write your answer in your own words. A good way to accomplish this is to close your books and browsers while you write up your solution.

For reference, here's the proof that the sum of the measures of the
exterior angles of a triangle is 360 degrees.

The sum m(a)+m(b)+m(c) of the measures of the interior angles of any triangle is 180.
Interior and exterior angles come in supplementary pairs: a and d, b
and e, c and f. So m(a)+m(d) = m(b)+m(e) = m(c)+m(f) = 180.

Solve for the measures of the exterior angles:

m(d) = 180 - m(a)

m(e) = 180 - m(b)

m(f) = 180 - m(c)

Sum the measures of the exterior angles:

m(d) + m(e) + m(f) = (180 - m(a)) + (180 - m(b)) + (180 - m(c))

m(d) + m(e) + m(f) = 180 + 180 + 180 - m(a) - m(b) - m(c)

m(d) + m(e) + m(f) = 180 + 180 + 180 - (m(a) + m(b) + m(c))

m(d) + m(e) + m(f) = 180 + 180 + 180 - (180)

m(d) + m(e) + m(f) = 360.

The sum of the measures of the exterior angles of a triangle is 360 degrees.