It is known that f is the point where AD edge pairs overlap on BC.

Then < AFE = 90.

∴∠ 1? +? ∠2? =90 ? (This need not be explained. )

Again? ∠C？ =? 90 ∴∠2? +? ∠3? =90 ?

Similarly ∠ 1? +? ∠4=90

∴∠ 1? =? ∠3∠2=∠4

∴? △ABF？ ∽? △FCE？

We also know that tan∠EFC=3/4.

tan∠EFC=CE/FC=3/4

Let CE=3x and FC=4x.

Then according to Pythagorean theorem, FE=5x is obtained.

∵DE=EF∴DC=DE+EC=8X？

According to the proportion of the corresponding side of similar triangles.

Get CE/BF? =? FC/AB

Namely. 3x/BF？ =4x/8x？

Get BF=6x.

∴? BC? =BF+FC= 10x

According to Pythagorean theorem

AE=AD+DE

Namely. (5 times the root number 5? )=? ( 10x)+(5x)？

The solution is x= 1.

∴? AD=BC= 10？ AB=DC=8

∴C？ =? ( 10+8)×2=36