Question

PQRS is a parallelogram. A and B are midpoints of the sides PQ and RS
respectively.Prove that the line segment PB and AR trisect the diagonal SQ.

Answer 1

$\sf \orange{ABCD \: is \: // \: gm }\\ \sf \orange{AB \: // \: CD} \\ \sf \orange{AE \: // \: FC} \\ \sf \orange{⇒ AB = CD} \\ \sf \orange{\frac{1}{2}AB =\frac{1}{2}CD} \\ \sf \orange{AE \: = \: EC} \\\sf \orange{∆ECF \: is \: // \: gm }\\\sf \orange{ In DQC} \\ \sf \orange{F \: is \: mid \: point \: of \: DC} \\\sf \orange{ FP \: // \: CQ} \\ \sf \orange{By \: converse \: of \: mid \: point \: theorem \: P \: is \: mid \:point \: of \: DQ} \\\sf \orange{ ⇒ DP \: = \: PQ \: (1)} \\ \sf \orange{AF \: and \: EC \: bisect \: BD} \\\sf \orange{In \: ΔΑΡΒ}\\ \sf \orange{E \: is \: mid \: point \: of \: AB} \\ \sf \orange{EQ \: // \: AP }\\ \sf \orange{By \: converse \: of \: MPT \: (mid \: point \: theorem)} \\\sf \orange{ Q \: is \: mid \: point \: of \: PB }\\ \sf \orange{⇒PQ \: = QB \: (2)} \\ \sf \orange{By \: (1) \: and \: (2)} \\ \sf \orange{⇒ PQ \: = \: QB \: = \: DP }\\\sf \orange{ AF \: and \: EC \: bisect \: BD.}$