Q.If $\overrightarrow{OA}=\bar a$ and $\overrightarrow{OB}=\bar b$ then show that the vector along the angle bisector of angle AOB is given by $\bar d=\lambda\left(\dfrac{\bar a}{|\bar a|}+\dfrac{\bar b}{|\bar b|}\right)$, where $\lambda$ is a real number.

Write each midpoint's position vector using the midpoint formula, then add the three vectors.

Let position vectors of $A,B,C$ be $\bar a,\bar b,\bar c$ (origin $O$). Since $D,E,F$ are midpoints of $BC,CA,AB$:

$\bar d=\frac{\bar b+\bar c}{2},\qquad \bar e=\frac{\bar c+\bar a}{2},\qquad \bar f=\frac{\bar a+\bar b}{2}$

$\overline{AD}=\bar d-\bar a=\frac{\bar b+\bar c}{2}-\bar a$

$\overline{BE}=\bar e-\bar b=\frac{\bar c+\bar a}{2}-\bar b$

$\overline{CF}=\bar f-\bar c=\frac{\bar a+\bar b}{2}-\bar c$

Adding: …

Use $A=-B$ (diametrically opposite about center $O$) and show $\vec{PA}\cdot\vec{PB}=0$.

Let $O$ be the centre, radius $r$, with $A,B$ the endpoints of a diameter, so $\vec{OA}=-\vec{OB}$ (i.e. $\bar a=-\bar b$). Let $P$ be any point on the circle, $\vec{OP}=\bar p$, with $|\bar p|=|\bar a|=|\bar b|=r$.

$\vec{PA}=\bar a-\bar p=-\bar b-\bar p$

$\vec{PB}=\bar b-\bar p$

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Show the point dividing each median in ratio $2:1$ from the vertex is the same point $\dfrac{\bar a+\bar b+\bar c}{3}$ for all three medians.

Let $\triangle ABC$ have position vectors $\bar a, \bar b, \bar c$ with respect to origin $O$.

Let $D$ be the midpoint of $BC$. Then

$\bar d = \frac{\bar b + \bar c}{2}$

Let $G$ be the point on median $AD$ dividing it internally in the ratio $2:1$ from $A$ (i.e. $AG:GD = 2:1$). By the section formula:

$\bar g = \frac{2\bar d + 1\cdot\bar a}{2+1} = \frac{2\left(\dfrac{\bar b+\bar c}{2}\right)+\bar a}{3} = \frac{\bar a+\bar b+\bar c}{3}$

Now let $E$ be the midpoint of $AC$, so $\bar e = \dfrac{\bar a+\bar c}{2}$. Let $G'$ divide median $BE$ in ratio $2:1$ from $B$:

$\bar{g'} = \frac{2\bar e + \bar b}{3} = \frac{2\left(\dfrac{\bar a+\bar c}{2}\right)+\bar b}{3} = \frac{\bar a+\bar b+\bar c}{3}$

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Show the point dividing each median in ratio $2:1$ from the vertex is the same point $\dfrac{\bar a+\bar b+\bar c}{3}$ for all three medians.

Let $\triangle ABC$ have position vectors $\bar a,\bar b,\bar c$ for its vertices A, B, C respectively.

Median from A: Let D be the midpoint of BC, so $\bar d=\dfrac{\bar b+\bar c}{2}$. Let G be the point on AD dividing it internally in ratio $2:1$ (from A):

$\bar g=\frac{2\bar d+1\cdot\bar a}{2+1}=\frac{2\left(\frac{\bar b+\bar c}{2}\right)+\bar a}{3}=\frac{\bar a+\bar b+\bar c}{3}$

Median from B: Let E be the midpoint of AC, $\bar e=\dfrac{\bar a+\bar c}{2}$. The point dividing BE in ratio $2:1$ from B is:

$\frac{2\bar e+\bar b}{3}=\frac{2\left(\frac{\bar a+\bar c}{2}\right)+\bar b}{3}=\frac{\bar a+\bar b+\bar c}{3}$

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