The correct option is A True
Given: △ABC, P is mid point of BC, QR∥BC and PQ∥AC
Since, PQ∥AC and P is mid point of BC, thus, by converse of mid point theorem
Q is mid point of AB.
Now, In △ABP
Since, QR∥BP and Q is mid point of AB. thus, by converse of Mid point theorem
R is mid point of AP.
Hence, AP=2AR