Question

ABCD is a parallelogram in which x and y are midpoints of Ab and CD respectively a y and DX are joined which intersect each other at p b y MCX are also join which intersect each other at Q show that PX Qy is a parallelogram

Answer 1

Given: X and Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD.
AX = XB and DY = YC
To Prove: PXQY is a parallelogram.
Proof: AB = DC 
... AB =   DC 
XB = DY......(i)
Also AB ïï DC...... (opp. sides of a ïïgm)
      XB ïï DY.....(ii)
Since in quadrilateral XBYD, XB = DY and XB ïï DY
... XBYD is a parallelogram.
... DX ïï YB ⇒ PX ïï YQ......(iii)
Similarly we can prove, PY ïï XQ ......(iv)
From (iii) and (iv), we get PXQY is a parallelogram. 

Answer 2

Answer:

Since X is the midpoint, AX = XB, Y is the midpoint, CY = YD.

ABCD is a parallelo gram, therefore, AB = CD.

Therefore, AX = CY

Also, AB is parallel to DC.

Therefore, AX is parallel to CY

Thus, AXCY is a parallelogram.

Consequently, PY is parallel to QX.

Since X is the midpoint, AX = XB, Y is the midpoint, CY = YD.

ABCD is a parallelo gram, therefore, AB = CD.

Therefore, BX = DY

Also, AB is parallel to DC.

Therefore, BX is parallel to DY

Thus, BXDY is a parallelogram.

Consequently, PX is parallel to QY.

Thus, the arguments are,

1. PY is parallel to QX

2. PX is parallel to QY

Therefore, PXQY is a parallelogram.

Hope it helps you