Question

WXYZ is a rhombus. A, B, C and D are the midpoints of sides WX, XY, YZ and WZ respectively. show that ABCD is a rectangle.​

Answer 1

Given:

WXYZ is a rhombus.

A, B, C, D are mid points of WX, XY, YZ, WZ respectively.

To show:

ABCD is a rectangle.

Construction:

Join XZ and WY

Sol:

In Triangle WXZ,

since A and D are mid points of side WX and WZ

therefore,by mid point theorem,

AD // XZ ....... (i)

and, AD = ½XZ ..........(ii)

Similarly, BC//XZ........(iii)

and, BC=½XZ.......(iv)

From (i) and (iii),

AD//BC..........(v)

From (ii) and (iv)

AD=BC...........(vi)

From (v) and (vi)

quadrilateral ABCD is a parallelogram.

Now, in quadrilateral AMON,

AN//MO(SINCE, AD//XZ)

AND, AM//ON(SINCE, AB//WY)

Therefore,AMON is a parallelogram.

and, angleNAM=angleNOM(opposite angle)

angle NAM=90° ( since diagonala of a rhombus

intersect at 90°l

there fore,angleDAB =90°

Hence, ABCD is a rectangle..............showed