Question

Let P,Q,R,S as a mid point of side AB,BC,CD,DA of quadrilateral ABCD. Show that ABCD is a parallelogram.​

Answer 1

$\bf\underline{\underline{\pink{Correct\:Question:-}}}$

★ Let P,Q,R,S as a mid point of side AB,BC,CD,DA of quadrilateral ABCD. Show that ABCD is a parallelogram.

ar(||gm PQRS) = ½ar(quad.ABCD)

$\bf\underline{\underline{\blue{Solution:-}}}$

Join AC to AR

In ∆ABC, P and Q are the midpoints of the sides AB and AC respectively.

∴ PQ||AC and PQ=½AC

In ∆DAC, S and R are the midpoints of the sides AD and DC respectively.

∴ SR||AC and SR = ½AC

Thus, PQ || SR and PQ = SR

∴ PQRS is a Parallelogram

Now, medians AR devides ∆ACD into two ∆ of equal area

∴ ar(∆ARD) = ½ar(∆ACD)

Medians RS devides ∆ARD into two ∆ of equal area

∴ ar(∆DSR) = ½ar(∆ARD)

From (i) and (ii), we get ar(∆DSR) = ¼ar(∆ABC)

Similarly, ar(∆BPQ) = ¼ar(∆ABC)

∴ ar(∆DSR) + ar(∆BPQ) = ¼[ar(∆ACD)+ar(∆ABC)]

==> ar(∆DSR)+ar(∆BPQ) = ¼ar(quad. ABCD)

Similarly, ar(∆CRQ) + ar;∆ASp) = ¼ar(quad. ABCD)

Adding (iii) and (iv), we get

ar(∆DSR)+ar(∆BPQ)+ar(∆CRQ)+ar(∆ASP)+ar(||gm PQRS) = ar(quad. ABCD)

Substracting from (v) and (vi), we get

ar(||gm PQRS) = ½ar(quad. ABCD)

Hence, proved.

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Answer 2

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