`square ABCD` is a parallelogram point E is on side BC. Line DE intersects ray AB in point `T`. Prove that `DExxBE=CExxTE`.

Explanation

`segAB||segCD`
…..(Opposite sides of parallelogram are parallel)
i.e. set `AT||` seg CD ……….`(A-B-T)`
Line TD in the transversal,
`:./_ATD~=/_CDT` …..(Alternate angles)
i.e `/_BTE~=/_CDE……..(A-B-T,T-E-D)`..1
In `DeltaBET` and `DeltaCED` .....[From 1]
`:.DeltaBET~DeltaC`........(Vertically opposite angles)
`:DeltaBET~DeltaCED` ........(AA test of similarity)
`:.(BE)/(CE)=(TE)/(DE)`....(Corresponding sides of similar triangles are in proportion)
`:.BExxDE=CExxTE`.