Show that the quadrilateral formed by joining the mid points of the sides of the square is Rhombus.
Answers
Answer:
Step-by-step explanation:
In △ABC
,
P is the midpoint of AB and Q is the midpoint of BC
Using mid-point theorem, we can write
⇒PQ∥AC
and PQ=12AC
………….… (1)
In △ADC
,
S is the midpoint of AD and R is the midpoint of CD
Using mid-point theorem, we can write
⇒RS∥AC
and RS=12AC
………...… (2)
From equations (1) and (2),
PQ∥RS
(As they both are parallel to the same line AC)
PQ=RS
(As they both are equal to (12AC)
∴PQ∥RS,PQ=RS
………...… (3)
In △ABD
,
P is midpoint of AB and S is mid-point of AD
Using mid-point theorem, we can write
⇒PS∥BD
and PS=12BD
………...… (4)
In △BCD
,
Q is midpoint of BC and R is midpoint of CD
Using mid-point theorem, we can write
⇒QR∥BD
and QR=12BD
…………….… (5)
From equations (4) and (5),
PS∥QR
(As they both are parallel to the same line BD)
PS=QR
(As they both are equal to (12BD)
∴PS∥QR,PS=QR
……………….… (6)
From equations (3) and (6),
PQ∥RS,PS∥QR
and PQ=RS,PS=QR
⇒PQRS
is a parallelogram
We know ABCD is a rectangle,
So, opposite sides of the rectangle ABCD must be equal to each other.
Let us consider the pair of opposite sides as AD and BC.
⇒AD=BC
Divide both sides by 2
⇒12AD=12BC
……………...… (7)
Since, S is mid-point of AD ⇒AS=SD=12AD
Similarly, Q is midpoint of BC ⇒BQ=QC=12BC
Substitute the value of 12AD=AS,12BC=BQ
in equation (7)
⇒AS=BQ
………...… (8)
Now we prove the two triangles, △PAS,△PBQ
congruent to each other.
In△PAS,△PBQ
,
(Since P is the midpoint of AB ⇒PA=PB=12AB
)
PA=PB
(As A and B are vertices of rectangle ABCD and rectangle has all its interior angles as 90∘
)
∠A=∠B
AS=BQ
(From equation (8)
△PAS≅△PBQ
by SAS congruence rule.
We know that the corresponding sides of congruent triangles are equal to each other.
⇒PS=PQ
………...… (9)
We know from equations (3) and (6) PQ=RS,PS=QR
Substitute the value of PS=PQ
from equation (9)
⇒PQ=QR=RS=SP
So, all sides of the parallelogram are equal in length.
∴PQRS
is a rhombus.
Note:
Students can make mistakes while proving the SAS congruence as they assume any two sides and any angle of one triangle equal to corresponding sides and angle of another triangle. Keep in mind we take two sides and the included angle i.e. the angle between the two sides of one triangle equal to two sides and included angle of the other triangle.
