In a parallelogram ABCD, P is the midpoint of AD. Also, BP and AC intersect at Q. Then AQ : QC =
Take A as the origin, with \(\vec{AB} = \vec{b}\) and \(\vec{AD} = \vec{d}\). Then C = \(\vec{b} + \vec{d}\) and P = \(\frac{\vec{d}}{2}\).
A point on BP: \(\vec{b} + t\left(\frac{\vec{d}}{2} - \vec{b}\right)\). A point on AC: \(s(\vec{b} + \vec{d})\).
Compare coefficients: \(1 - t = s\) and \(\frac{t}{2} = s\). So \(s = \frac{1}{3}\).
So AQ = \(\frac{1}{3}\)AC, and AQ : QC = 1 : 2.
Answer: (D) 1 : 2
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