Question

A,B,C and D are four points on the sides PQ,QR,RS and SP respectively of a parallelogram PQRS such that PA=QB=RC=SD. Prove that ABCD is a parallelogram

Answer 1

Draw the diagram first:


Using the mid-point theorem on triangles △PQR△PQR and △PSR△PSR, it is clear that AB∥PR∥CDAB∥PR∥CD and AB=CD=PR2AB=CD=PR2. QED.

Update: It seems OP rephrased the question to the following (see comments):

PQRS is a quadrilateral. A, B, C, D are the midpoints of sides PQ, QR, RS and SP respectively and are joined to form a parallelogram. Show that area of parallelogram ABCD is half the area of quadrilateral PQRS. ( asked in this year's 9th std final paper.)

So here is modification to my original answer as follow up:

Note that, in terms of area,

△PAD=14△PQS△PAD=14△PQS.
△QAB=14△QP△QAB=14△QPR.
△RBC=14△RQS△RBC=14△RQS.
△SDC=14△SPR△SDC=14△SPR.
Adding the four equations, we get:

△PAD+△QAB+△RBC+△SDC=14(△PQS+△QPR+△RQS+△SPR)△PAD+△QAB+△RBC+△SDC=14(△PQS+△QPR+△RQS+△SPR)

Note that △PQS+△QPR+△RQS+△SPR=2×△PQS+△QPR+△RQS+△SPR=2× area of quadrilateral PQRSPQRS.

Whence, △PAD+△QAB+△RBC+△SDC△PAD+△QAB+△RBC+△SDC is half the area of PQRSPQRS. But, area of PQRS−△PAD+△QAB+△RBC+△SDCPQRS−△PAD+△QAB+△RBC+△SDC also equals the area of parallelogram ABCDABCD.

Answer 2

Answer:

i also want same answer

Step-by-step explanation:

someone pls help