Question

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 cm 

B. 3.5 cm 

C. 2.5 cm 

D. 5 cm

Answer 1

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