Option : (B)
Complete the parallelogram ADCP
So, the diagonals DP & AC bisect each other at O
Thus,
O is the midpoint of AC as well as DP ...(i)
Since ADCP is a parallelogram,
AP = DC
And,
AP parallel DC
But,
D is mid-point of BC (Given)
AP = BD
And,
AP parallel BD
Hence,
BDPA is also a parallelogram.
So,
Diagonals AD & BP bisect each other at E
(E being given mid-point of AD)
So,
BEP is a single straight line intersecting AC at F
In triangle ADP,
E is the mid-point of AD and
O is the midpoint of PD.
Thus,
These two medians of triangle ADP intersect at F,
which is centroid of triangle ADP.
By property of centroid of triangles,
It lies at \(\frac{2}{3}\) of the median from vertex
So,
AF = \(\frac{2}{3}\)AO ...(ii)
So,
From (i) and (ii),
AF = \(\frac{2}{3}\) x \(\frac{1}{2}\) x AC
= \(\frac{1}{3}\)AC
= \(\frac{10.5}{3}\)
= 3.5 cm