If D- E are the mid-points of AB- AC of triangle ABC and overline - DE - -lambda overline - BC - then lambda -12dfrac-1-2-3

Question

Toppr Admin · Accepted Answer

By mid point theorem- if D and E are mid points of AB and AC- then-xA0-xA0-xAF-DE-12-xAF-BC - -1-As per question -xAF-DE-x3BB-xAF-BC - -2-From -1- and -2- comparing- we get-xA0-x3BB-1-2So- option -C- is correct-

If $D, E$ are the mid-points of $A B$ , $A C$ of $△ A B C$ and $¯ ¯¯¯¯¯¯¯ ¯ D E = λ ¯ ¯¯¯¯¯¯ ¯ B C$ then $λ$ =
$1$
$2$
$\frac{1}{2}$
$3$

By mid point theorem, if $D$ and $E$ are mid points of $A B$ and $A C$ , then
$¯ D E = \frac{1}{2} ¯ B C$ ------ $(1)$
As per question $¯ D E = λ ¯ B C$ -------- $(2)$
From $(1)$ and $(2)$ , comparing, we get
$λ = 1 / 2$
So, option $(C)$ is correct.

If D,E are the mid-points of AB, AC of △ABC and ¯¯¯¯¯¯¯¯¯DE=λ¯¯¯¯¯¯¯¯BC then λ =12123

By mid point theorem, if D and E are mid points of AB and AC, then ¯DE=12¯BC ------ (1)As per question ¯DE=λ¯BC -------- (2)From (1) and (2), comparing, we get λ=1/2So, option (C) is correct.

If $D, E$ are the mid-points of $A B$ , $A C$ of $△ A B C$ and $¯ ¯¯¯¯¯¯¯ ¯ D E = λ ¯ ¯¯¯¯¯¯ ¯ B C$ then $λ$ =
$1$
$2$
$\frac{1}{2}$
$3$

By mid point theorem, if $D$ and $E$ are mid points of $A B$ and $A C$ , then
$¯ D E = \frac{1}{2} ¯ B C$ ------ $(1)$
As per question $¯ D E = λ ¯ B C$ -------- $(2)$
From $(1)$ and $(2)$ , comparing, we get
$λ = 1 / 2$
So, option $(C)$ is correct.