If D is the midpoint of the side Bc of a triangle ABC, then prove that $\to A B + \to A C = \to 2 A D$

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$\begin{matrix} P o i n t D i s t h e m i d - p o i n t o f B C s o, b y sec t i o n f o r m u l a \frac{¯ ¯¯¯¯¯ ¯ A B + ¯ ¯¯¯¯¯ ¯ A C}{2} = ¯ ¯¯¯¯¯¯¯ ¯ A D ¯ ¯¯¯¯¯¯ ¯ A B + ¯ ¯¯¯¯¯¯ ¯ A C = 2 ¯ ¯¯¯¯¯¯¯ ¯ A D \end{matrix}$

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If D is the midpoint of the side Bc of a triangle ABC, then prove that →AB+→AC=→2AD

PointDisthemid−pointofBCso,bysectionformula¯¯¯¯¯¯¯AB+¯¯¯¯¯¯¯AC2=¯¯¯¯¯¯¯¯¯AD¯¯¯¯¯¯¯¯AB+¯¯¯¯¯¯¯¯AC=2¯¯¯¯¯¯¯¯¯AD

If D is the midpoint of the side Bc of a triangle ABC, then prove that $\to A B + \to A C = \to 2 A D$

$\begin{matrix} P o i n t D i s t h e m i d - p o i n t o f B C s o, b y sec t i o n f o r m u l a \frac{¯ ¯¯¯¯¯ ¯ A B + ¯ ¯¯¯¯¯ ¯ A C}{2} = ¯ ¯¯¯¯¯¯¯ ¯ A D ¯ ¯¯¯¯¯¯ ¯ A B + ¯ ¯¯¯¯¯¯ ¯ A C = 2 ¯ ¯¯¯¯¯¯¯ ¯ A D \end{matrix}$