To prove that ar(triangle ABE) = ar(trapezium ABCD)

[elephantsize]

:?: :?: :?: :?: :?: :?: :?: :?:

:shock: In a trapezium ABCD, the diagonal AC is constructed. From D, a line segment DE is constructed parallel to AC meeting BC produced to K in E. AE is joined. Prove that the newly formed triangle ABE is equal in area to the trapezium ABCD. :shock:

 

[soroban]

Hello, elephantsize!

Is there a typo in the problem? \(\displaystyle \;\)As stated, it is not true.

In trapezium ABCD, the diagonal AC is constructed. \(\displaystyle \;\)From D, a line segment DE is constructed
parallel to AC meeting BC produced to K in E. \(\displaystyle \;\)AE is joined.
Prove that the newly formed triangle ABE is equal in area to the trapezium ABCD.

          A          B
          *----------*
         /   *        \
        /       *      \
       /           *    \
      /                * \
    D*-----------+--------*C
         *       F         \
            *               \
               *             \
                  *           \
                     *         \
                        *       \
                           *     \
                              *   \
                                 * \
                                    *
                                     E

Draw AE, intersecting DC in F.

It is claimed that the area of triangle ABE = area of trapezoid ABCD.

Both shapes share the quadrilateral ABCF.

Hence, \(\displaystyle \Delta FCE\,=\,\Delta ADF\) . . . which is highly unlikely.