Question

X and Y are respectively the mid point of sides AB and BC of a parallelogram ABCD.DX and DY intersect AC at M and N respectively. If AC=4.5cm find MN.

Answer 1

Solution:-

ABCD is a parallelogram. X and Y are the mid points of the sides AB and BC respectively. DX and DY intersect AC at M and N respectively.
In Δ ABC, X and Y are mid points of AB and BC respectively.
∴ XY = 1/2AC and XY || AC .........(1)               (Mid point theorem)
In Δ AOB, 
X is the mid point of AB and XS || OM  (XY || AC)
∴ S is the mid point of OB  (Converse of mid point theorem 
⇒ OS = OB
We know that, diagonals of the parallelogram bisect each other.
∴ OD = OB
⇒ OD = OS + SB
⇒ OD = 2OS  (OS = SB)
⇒ OD/OS = 2
⇒ OS/OD = 1/2
⇒ OS/OD + 1 = 1/2 + 1
⇒ (OS + OD)/OD = 3/2
⇒ DS/OD = 3/2
⇒ OD/DS = 2/3
Δ DMO ~ Δ DNO    (AA similarity)
∴ MO/XS = OD/DS 
⇒ MO/XS = 2/3
⇒ MO = 2/3XS  ..............(2)
Similarly ON = 2/3SY  ...........(3)
Adding (2) and (3), we get.
MO + ON = 2/3(XS + SY)
∴ MN = 2/3XY
⇒ MN = 2/3 × 1/2AC   (Using equation 1)
⇒ MN = 2/3 × 4.5/2
⇒ MN = 9/6
⇒ MN = 1.5 cm
Answer.