Question

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

C. ABCD is a Rhombus

D. Diagonals of ABCD are perpendicular

Answer 1

ABCD is a quadrilateral, M, N, O,P are the midpoints of of sides of AB, BC, CD, and DA,

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 = 1/2 DB .........(i)

same in triangle ABC,

MN II AC

MN = 1/2 AC ( using the midpoint theorem)...........(ii)

from equation (i)

PM =MN

1/2 DB = 1/2 AC

DB = AC

2)hence diagonals of quadrilatera are equal, and

3)quadrilateral ABCD is a square not a rhombus, and

4)diagonals are also perpendicular,