If C is the mid point of AB and P is any point outside AB, then

$\vec{PA}+\vec{PB}=2\vec{PC}$

Using triangle law of addition of vectors in triangles PAC and PBC, we have

$\vec{PA}+\vec{AC}=\vec{PC}$ and $\vec{PB}+\vec{BC}=\vec{PC}$

$⇒\vec{PA}+\vec{AC}+ \vec{PB}+\vec{BC}=\vec{PC} + \vec{PC}$

$⇒\vec{PA}+\vec{PB}+(\vec{AC}+\vec{BC}) = 2\vec{PC}$

$⇒\vec{PA}+\vec{PB}+(\vec{AC}-\vec{AC})=2\vec{PC}$

$⇒\vec{PA}+\vec{PB}=2\vec{PC}$