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The figure formed by joining the mid-points
of the sides of a quadrilateral ABCD, taken in
order, is a square only, if
A. Diagonals of ABCD are equal
B. Diagonals of ABCD are equal and perpendicular
D. Diagonals of ABCD are perpendicular
Given
a quadrilateral ABCD where M, N, O, P are the midpoints of sides of AB, BC, CD, and DA
To prove MNOP is a square if
A. Diagonals of ABCD are equal
B. Diagonals of ABCD are equal and perpendicular
C. ABCD is a Rhombus
D. Diagonals of ABCD are perpendicular
Solution:
ABCD is a quadrilateral, Let M, N, O, P are the midpoints of sides of AB, BC, CD, and DA of quadrilateral ABCD
therefore:---
MNOP is a square,
where MN = NO = OP = PN
and MO = NP , PN = AB
1) Thus all sides of a quadrilateral are equal.
therefore, quadrilateral ABCD is either a square or a rhombus,
Now in triangle ADB, using the midpoint theorem
PM II DB,
therefore, PM = $\frac{1}{2}$ DB ...(i)
same in triangle ABC,
MN II AC
MN = $\frac{1}{2} AC using the midpoint theorem....(ii)
from equation (i)
PM =MN
$\frac{1}{2} DB = $\frac{1}{2}$ AC
DB = AC
2) Hence diagonals of the quadrilateral are equal, and
3)quadrilateral ABCD is a square, not a rhombus, and
4)diagonals are also perpendicular.
Updated on: 10-Oct-2022
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