[[1]]is formed when the midpoints of the pairs of adjacent sides of a rhombus are joined.

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Accepted Answer

Rectangle is formed when the midpoints of the pairs of adjacent sides of a rhombus are joined.Let us draw the rhombus ABCD (say). Joining the midpoints E, F, G, and H as shown in the figure.Consider

ΔAEF
   and

ΔGHC
  .We know that all sides of a rhombus are equal in length, and also that the opposite angles are equal, therefore, AD = BC and

1
    2
   
   AD=
    1
    2
   
   BC
  , (using midpoint theorem) AB = CD and

1
    2
   
   AB=
    1
    2
   
   CD
  . (Using midpoint theorem)Consider

ΔAEF
   and

ΔGHC
   ,AE = GC

∠FAE=∠GCH
   AF = CHTherefore, by SAS criterion of congruence,

ΔAEF≅ΔBHC
  . EF = GH         … (By C.P.C.T)                              Similarly,

ΔDEH≅ΔBFG
  . EH = FG         … (By C.P.C.T)                                                        In

ΔEDH
  , ED = DH and it is an isosceles triangle.

∠DEH=∠DHE=x

∠DEH+∠DHE+∠HDE=
      180
      °

2x+∠HDE=
      180
      °

x=

180
      °
     
     −∠HDE
    2

In

ΔAEF
  , AE = AF and it is an isosceles triangle.

∠AEF=∠AFE=y

∠AEF+∠AFE+∠FAE=
      180
      °

2y+∠FAE=
      180
      °

y=

180
      °
     
     −∠FAE
    2

We know that the adjacent angles of a rhombus are supplementary.

y=

180
      °
     
     −(
      
       180
      °
     
     −∠HDE)
    2

y=
    
     ∠HDE
    2

Substituting the values of x and y from above equations we get,

x+y+∠FEH=
      180
      °

180
        °
       
       −∠HDE
      2
     
     +
      
       ∠HDE
      2
     
     +∠FEH=
      180
      °

90
      °
     
     +∠FEH=
      180
      °

∠FEH=
      90
      °

Similarly, we can show that

∠EHG,∠FGH,∠EFG
   are all

90
    °

. Thus, EFGH is a rectangle. We can conclude that a rectangle is formed when the midpoints of the pairs of adjacent sides of a rhombus are joined.

[[1]]is formed when the midpoints of the pairs of adjacent sides of a rhombus are joined.

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