In general a trapezoid is defined as a quadrilateral with two parallel sides. An isosceles trapezoid is a special case of trapezoid which has lateral symmetry, meaning that one side would be a mirror of the other.
The image above depicts an isosceles trapezoid, #"ABCD"#. It should be apparent that triangles #"ABE"# and #"DCE"# have equal areas, as they are mirror images of each other. Therefore, lets focus on trying to prove that these two triangles should have the same area for all trapezoids.
The area of a triangle can be found using the identity:
#A = 1/2 (B xx H)#
Where #B# is the base of the triangle, and #H# is the height. Triangles #"ABE"# and #"DCE"# do not share any sides that we could consider a shared base, however, triangles #"ACD"# and #"DBA"# do share a common side, the base of the trapezoid.
If we consider side #"AD"# the base for both triangles, then the height for both is the distance between line #"AD"# and #"BC"#. Since the base and height are equal for both triangles, then the area for both must also be equal.
#A_"ABD" = A_"DCA"#
Now we can also see that triangle #"ABD"# is composed of triangles #"ABE"# and #"AED"#. Furthermore, triangle #"DCA"# is composed of triangles #"DCE"# and #"AED"#.
#A_"ABD" = A_"ABE" + A_"AED"#
#A_"DCA" = A_"DCE" + A_"AED"#
If we subtract the area of AED from both ABD and DCA, we can see that the areas of ABE and DCE are indeed equal.
#A_"ABD" = A_"DCA"#
#A_"ABE" + color(red)cancel(color(black)(A_"AED")) = A_"DCE" + color(red)cancel(color(black)(A_"AED"))#
#A_"ABE" = A_"DCE"#
Although I illustrated this proof with an isosceles trapezoid, none of the work was specific to a regular trapezoid. Therefore, the side triangles for any trapezoid can be shown to be equal using the same reasoning.
