L, M,K,N,K are mid points of sides BC , CD , DA , and AB respectively of square ABCD , prove that DL,DK,BM, and BAN enclose a rhombus?
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Proof In AADC, S and R are the mid-points of AD and DC respectively. Then, by mid-point theorem.
SR || AC and SR= 1 2 AC
...(0)
In AABC, P and Q are the mid-points of AB and BC respectively. Then, by mid-point theorem
PQ|| AC and PQ AC = 2 1 SR=PQ= AC =
...(ii)
From Eqs. (i) and (ii),
. (iii)
Similarly, in ABCD,
RQ|| BD and RQ= BD
...(iv)
And in ABAD,
1 SP || BD and SP = BD 2
...(v)
From Eqs. (iv) and (v).
SP = RQ = 1 BD = 2 2 AC
[given, AC BD]...(vi)
From Eqs. (iii) and (vi), SRPQ=SP = RQ It shows that all sides of a quadrilateral PQRS are equal. Hence, PQRS is a rhombus.
Hence proved.
Step-by-step explanation:
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