Question

State true or false:

$D,E$ and $F$ are the mid-points of the sides $AB,BC$ and $CA$ respectively of $△ABC$. $AE$ meets $DF$ at $O$. $P$ and $Q$ are the mid-points of $OB$ and $OC$ respectively. Then $DPQF$ is a parallelogram.

• A. True
• B. False

Answer 1

The correct option is A True

Given: $△ABC$, D, E, F are mid points of AB, BC, AC respectively.
AB and DF meet at O. P and Q are mid points of OB and OC respectively.
To Prove: DPFQ is a parallelogram.
Now, In $△ABC$
D is mid point of AB and F is mid point of AC
Hence, $DF\parallel BC$ and $DF=\frac{1}{2}BC$... (1) (Mid point theorem)
In $△OBC$
P is mid point of OB and Q is mid point of OC
Hence, $PQ\parallel BC$ and $PQ=\frac{1}{2}BC$... (2) (mid point theorem)
thus, from (1) and (2)
$DF\parallel PQ$ and $DF=PQ$ (5)
Now, In $△ABE$
D is mid point of AB and P is mid point of OB
Thus, $DP\parallel AE$ and $DP=\frac{1}{2}AE$ (3) (mid point theorem)
Now, In $△ACE$
F is mid point of AC and Q is mid point of OC
Thus, $FQ\parallel AE$ and $QF=\frac{1}{2}AE$ (4) (mid point theorem)
thus, from (3) and (4)
$DP\parallel FQ$ and $DP=FQ$ (6)
Now, from (5) and (6)
DPQF is a parallelogram