In Δ ABC, E is the mid-point of median AD such that BE produced meets

1.	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» 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

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»

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