Consider triangles BEF and CDF,
#BP = CP " given"#
#hat (BPE) = hat (CPD), " Vert. opp. angles"#
# hat (PBE) = hat (DPC) " Alt. angles as " vec(BE) " // " vec (DC)#
Hence #color(crimson)(Delta BEP " congruent " Delta CDP, " ASA theorem"#
#:. bar(BE) = bar (DC)#
#color(blue)(bar (AB) = bar (DC), " opp. sides of ABCD parallelogram"#
#"Likewise", color(blue)(bar (DC) = bar (EF), " opp. sides of CDEF parallelogram"#
#"i.e. "color(green)( bar (AB) = bar (BE) = bar (EF) = bar (DC)#
#color(maroon)(bar (AF) = bar (AB) + bar (BE) + bar (EF) = 3 * bar(DC)#