Question

A, B, C, D are the mid point of the sides PQ, QR, RS, SP of a parallelogram PQRS respectively. SA, SB, QC and QD have been joined to intersect at E and F. Show that EQFS is a parallelogram
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Answer 1

Answer:

Step-by-step explanation:

Refer to the diagram in order to understand the question.

A & C are midpoints of PQ and SR respectively

Therefore,

AP = AQ = PQ/2 ....(i)

SC = CR = SR/2 ....(ii)

And, PQ = SR ...(iii) [opposite sides of a parallelogram are equal]

Therefore, from (i), (ii), and (iii),

AP = AQ = SC = CR ...(iv)

Now in ∆s APS & CRQ,

AP = CR [using (iv)]

Angle APS = Angle CRQ [opposite angles of a ||gm are equal]

PS = RQ [opposite sides of a ||gm are equal]

Therefore, ∆APS is congruent to ∆CRQ (via SAS Criterion)

=> AS = CQ [C.P.C.T.]

Since, AS = CQ & SC = AQ

=> ASCQ is a ||gm [opposite sides are equal]

Similarly, DSBQ is a ||gm

Since ASCQ is a ||gm

So, AS || CQ

=> ES || FQ ...(v) [opposite sides of a ||gm are parallel]

Also, DSBQ is a ||gm

So, DQ || SB

=> EQ || SF ...(vi) [opposite sides of a ||gm are parallel]

So, from (v) and (vi),

ESFQ is a ||gm

Hence, proved.