ABCD is a parallelogram and E and F are the centroids of triangles ABD

1.	ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF = A. AE B. BE C. CE D. DE
Answer» Option : (A) Given : ABCD is a parallelogram E and F are the centroids of triangle ABD and BCD Since, The diagonals of parallelogram bisect each other AO is the median of triangle ABD And, CO is the median of triangle CBD EO = \(\frac{1}{3}\)AO (Since, centroid divides the median in the ratio 2:1) Similarly, FO = \(\frac{1}{3}\)CO EO + FO = \(\frac{1}{3}\)AO +\(\frac{1}{3}\)CO = \(\frac{1}{3}\)(AO + CO) EF = \(\frac{1}{3}\)AC AE = \(\frac{1}{3}\)AO = \(\frac{2}{3}\) x \(\frac{1}{2}\) AC = \(\frac{1}{3}\)AC Therefore, EF = AE

ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF = A. AE B. BE C. CE D. DE

Answer»

Option : (A)

Given :

ABCD is a parallelogram

E and F are the centroids of triangle ABD and BCD

Since,

The diagonals of parallelogram bisect each other

AO is the median of triangle ABD

And,

CO is the median of triangle CBD

EO = \(\frac{1}{3}\)AO

(Since, centroid divides the median in the ratio 2:1)

Similarly,

FO = \(\frac{1}{3}\)CO

EO + FO = \(\frac{1}{3}\)AO +\(\frac{1}{3}\)CO

= \(\frac{1}{3}\)(AO + CO)

EF = \(\frac{1}{3}\)AC

AE = \(\frac{1}{3}\)AO

= \(\frac{2}{3}\) x \(\frac{1}{2}\) AC

= \(\frac{1}{3}\)AC

Therefore,

EF = AE