Question

EFGH is a rectangle and L, M, N and O are mid-points of the sides EF, FG, GH and HE respectively. Show that the quadrilateral LMNO is a rhombus.​

Answer 1

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Given : ABCD is a square. E,F,G and H are the midpoint AB,BC,CD and DA respectively. Such that AE=BF=CG=DH

To show : EFGH is a square

Proof :

AE=BF=CG=DH

Therefore,

BE=CF=DG=AH

In trinagle AEH and triangle BFE

AE=BF (Given)

∠A=∠B (each equal to 90

∘

)

AH=BE

By SAS criterion of congurency, triangle AEH is congruent to triangle BFE.

EH=BF (By CPCT)

Similarity,

EH=HG=GF=FE

Now,

∠AEH=∠BFE and ∠AHE=∠BEF

But, ∠AEH+∠AHE=90

∘

and ∠BFE+∠BFE=90

∘

∠AEH+∠AHE+∠BFE+∠BFE=90

∘

+90

∘

∠AEH+∠BEF+∠AEH+∠BEF=90

∘

+90

∘

2(∠AEH+∠BEF)=180

∘

∠AEH+∠BEF=90

∘

∠HEF=90

∘

Similarity,

∠EFG=∠FGH=∠GHE=90

∘

Therefore, EFGH is a square

Answer 2

Answer:

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