Question

In figure AD and BE are the medians of triangle ABC and BE IS Parallel toDF .prove that CF=1/4AC

Answer 1

Answer:

Consider ΔABC

Since AD is Median ⟹BD=BC⋯(1)

Since BE is Median ⟹AE=EC⋯(2)

Consider ΔBEC

Since BE∣∣DF

By Basic Proportionality theorm,

⟹

DC

BD

=

FC

EF

From (1)

⟹CF=EF

⟹CF=

2

1

AE

From (2)

⟹CF=

4

1

AC

Answer 2

$\pink\bigstar$Given:

❖ AD and BE are the medians of ∆ABC

❖ BE II DF

$\blue\bigstar$To Prove :

❖ CF = 1/4 AC

$\green\bigstar$Proof:

In ∆ABC,

$\because$ AD is the median of BC,

$\implies$ BD = BC ___ (i)

$\because$ BE is the median of AC,

$\implies$ AE = CE ___ (ii)

In ∆BEC,

$\because$ BE II DF and D is the mid - point of BC,

$\implies$ F is the mid - point of EC

[By converse of Mid-Point Theorem]

$\implies$ EF = CF

Now,

CF = $\sf\dfrac{EC} {2}$

$\implies$ CF = 1/2 ( 1/2 of AC)

[$\because$ EC = EA from equation (ii)]

$\implies$ CF = 1/4 AC

$\red\bigstar$ Concepts Used:

❖ Property of a median.

❖ Converse of Mid - Point Theorem.