Question

$\huge\orange {\boxed {\boxed {\mathtt {Question}}}}$

Show that the quadrilateral formed by joining the mid points of the adjacent side of a square is also a square.​

Answer 1

Answer:

Step-by-step explanation:

Let a square ABCD in which L,M,N&O are the midpoints .

in triangle AML and triangle CNO  

      AM = CN ( AB = DC and M and O are the midpoints )

      AL = CM ( AD = BC and L and   N are the midpoints )

      angle MAL = angle NCO ( all angles of a square = 90 degree )

     by AAS criteria

       triangle AML CONGRUENT to triangle CNO

   therefore ML = ON  ( CPCT  )

similarly in triangle MBN CONGRUENT to  LDO  and  

   AND triangle  AML is CONGRUENT  to triangle

now ,  

 in Triangle AML ,

angle AML = angle ALM ( AM = AL )  

                  = 45 degree

  similarly in triangle LDO  

  angle DLO = 45 degree

there fore ,

angle MLO = 90 degree  

by the properties of SQUARE  

all sides are equal and angles are 90 degree

Answer 2

${\huge \underline {\boxed {\bold {Question}}}}$

• Show that the quadrilateral formed by joining the mid points of the adjacent side of a square is also a square.

${\huge \underline {\boxed {\bold {Answer}}}}$

1) Consider △ ABC

We know that E and F are the midpoints

• Based on the midpoint theorem

We know that EF || AC and EF = ½ AC

2) Consider △ ADC

We know that H and G are the midpoints

• Based on the midpoint theorem

We know that HG || AC and HG = ½ AC

So we get EF || HG and EF = HG = ½ AC ……….. (1)

3) Consider △ BAD,

We know that H and E are the midpoints

• Based on the midpoint theorem

We know that HE || BD and HE = ½ BD

4) Consider △ BCD,

We know that G and F are the midpoints

• Based on the midpoint theorem

We know that GF || BD and GF = ½ BD

So we get HE || GF and HE = GF = ½ BD …….. (2)

We know that the diagonals of a square are equal

So we get AC = BD …….. (3)

By using equations (1), (2) and (3) ,

So we get GF || BD and HE || GF

We have EF = GH = GH = HE

We know that EFGH is a rhombus .

From the figure we know that the diagonals of a square are equal and intersect at right angles

∠ DOC = 90°

We know that the sum of adjacent angles of parallelogram is 180°

It can be written as ∠ DOC + ∠ GKO = 180°

By substituting the values

90° + ∠ GKO = 180°

On further calculation ∠ GKO = 180° – 90°

By subtraction ∠ GKO = 90°

From the figure we know that ∠ GKO and ∠ EFG are corresponding angles

We get ∠ GKO = ∠ EFG = 90°

We know that ∠ EFG = 90°

Hence, EFGH is a square.

Therefore, it is proved that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.