Question

5. A B C D are the mid-points of the sides PQ, QR, RS and SP of
a parallelogram PQRS respectively, SA, SB, QC and QD have been
jomed to intersect at E and f. Show that EQFS is a parallelogram.
[HOT]​

Answer 1

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.

Answer 2

Answer:

op

Step-by-step explanation: