In a triangle $A B C, D, E$ and $F$ are the mid-points of side $B C, C A$ and $A B$ respectively. Show that $¯ ¯¯¯¯¯¯¯ ¯ A D + ¯ ¯¯¯¯¯¯ ¯ B E + ¯ ¯¯¯¯¯¯ ¯ C F = ¯ ¯ ¯ 0$ .

Open in App

Solution

Let, $\to a, \to b, \to c$ represents
$A, B$ , and $C$ respectively
$∴ D = \frac{1}{2} (\to b + \to c)$
$E = \frac{1}{2} (\to a + \to c)$
$F = \frac{1}{2} (\to a + \to b)$
$∴ A D = \frac{1}{2} (v e c b + \to c) - \to a$
$B E = \frac{1}{2} (\to a + \to c) - \to b$
$C F = \frac{1}{2} (\to a + \to b) - \to c$
$∴ \to A D + \to B E + \to C F = \frac{1}{2} (\to b + \to c) - \to a + \frac{1}{2} (\to h + \to c) - \to b + \frac{1}{2} (\to a + \to b) - \to c$
$= \frac{1}{2} (2 \to b + 2 \to c + 2 \to a) - (\to a + \to b + \to c)$
$= (\to a + \to b + \to c) - (\to a + \to b + \to c)$
$= \to 0$ .

Suggest Corrections

Similar questions

A B C, D, E, F

B C, C A

A B

\to A D = - (\to B E + \to C F)

\vec{A D} + \vec{B E} + \vec{C F} .

A B C

D, E

F

B C, C A

A B

\frac{A B - B C}{2} < A E < \frac{A B + B C}{2} .

Δ A B C

\frac{(A D + B E + C F)}{(A B + B C + C A)}

∆ A B C, then \vec{A D} + \vec{B E} + \vec{C F} =______________________.

Join BYJU'S Learning Program