If C is the midpoint of AB and P is any point | vedclass

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(B) Since $C$ is the midpoint of $AB$,we have $\overrightarrow{AC} + \overrightarrow{BC} = 0$,which implies $\overrightarrow{AC} = -\overrightarrow{BC

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

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(B) Since $C$ is the midpoint of $AB$,we have $\overrightarrow{AC} + \overrightarrow{BC} = 0$,which implies $\overrightarrow{AC} = -\overrightarrow{BC}$ or $\overrightarrow{AC} = \overrightarrow{CB}$.Using the triangle law of vector addition in $	riangle PAC$ and $	riangle PBC$:$\overrightarrow{PA} = \overrightarrow{PC} + \overrightarrow{CA}$$\overrightarrow{PB} = \overrightarrow{PC} + \overrightarrow{CB}$Adding these two equations:$\overrightarrow{PA} + \overrightarrow{PB} = (\overrightarrow{PC} + \overrightarrow{CA}) + (\overrightarrow{PC} + \overrightarrow{CB})$$\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC} + (\overrightarrow{CA} + \overrightarrow{CB})$Since $C$ is the midpoint of $AB$,$\overrightarrow{CA} = -\overrightarrow{CB}$,so $\overrightarrow{CA} + \overrightarrow{CB} = 0$.Therefore,$\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC}$.

If $C$ is the midpoint of $AB$ and $P$ is any point outside $AB$,then:

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