Q. 325.0( 1 Vote )

# In Δ ABC, E is the mid-point of median AD such that BE produced meets AC at F. If AC = 10.5 cm, then AF =A. 3 cmB. 3.5 cmC. 2.5 cmD. 5 cm

So the diagonals DP & AC bisect each other at O

Thus O is the midpoint of AC as well as DP (i)

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

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 of the median from vertex

So,

AF = AO (ii)

So,

From (i) and (ii),

AF = * * AC

= AC

= = 3.5 cm

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