Triangles

construction: from E, draw EY parallel   to AC meeting BC in Y.

In triangle ADC,

EY II AC

So,

DE/EA=DY/YC...(i).. by BPT

but DE/EA=1... given

 so

 DY/YC=1..(ii)

so DC=2YC

now,

 in triangle XBC

EY II XC, since EY is parallel to AC by construction.

so,

BE/EX=BY/YC=(BD+DY)/YC=(DC+YC)/YC=(2YC+YC)/YC=3... since D is mid pt of BC.

Hence proved.