Show that the diagonals of a trapezoid divide it into four triangles of which two have the same area? "equi-extension"

Question

Show that the diagonals of a trapezoid divide it into four triangles of which two have the same area?  "equi-extension"

Aurora Ayala · Accepted Answer

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.

Parker Allen · Answer

undefined To show that the diagonals of a trapezoid divide it into four triangles, two of which have the same area, we can use the concept of equi-extension.

Let's consider a trapezoid with bases $ b_1 $ and $ b_2 $ and height $ h $. We draw its diagonals, dividing the trapezoid into four triangles: $ \triangle ADC $, $ \triangle ABC $, $ \triangle ABD $, and $ \triangle BCD $.

Now, let's denote the length of the diagonals as $ d_1 $ and $ d_2 $.

Using the concept of equi-extension, we can observe that the triangles $ \triangle ABD $ and $ \triangle BCD $ have the same area because they share the same base (segment $ BD $) and have equal heights.

Similarly, triangles $ \triangle ADC $ and $ \triangle ABC $ have the same area because they share the same base (segment $ AC $) and have equal heights.

Therefore, the diagonals of a trapezoid divide it into four triangles, two of which ( $ \triangle ABD $ and $ \triangle BCD $, $ \triangle ADC $ and $ \triangle ABC $) have the same area.