91Ó°ÊÓ

Consider a trapezoid, OABC, with OA parallel to CB.

Let PQ be the median of trapezoid.

Let $\stackrel{→}{OA}=\stackrel{→}{A}$ and $\stackrel{→}{OC}=\stackrel{→}{C}$

Since, $BCâˆ\yen OA$ so $\stackrel{→}{CB}=m\stackrel{→}{A}$

Now, using vector laws of addition

$$\begin{gathered} \stackrel{→}{OB}=\stackrel{→}{OC}+\stackrel{→}{CB} \\ =\stackrel{→}{C}+m\stackrel{→}{A} \end{gathered}$$

Since P is mid point of OC

So $\stackrel{→}{OP}=\frac{1}{2}\stackrel{→}{C}$

And Q is the mid point AB

So $\stackrel{→}{AQ}=\frac{1}{2}\stackrel{→}{AB}$

Now, again using Triangle law,

$$\begin{gathered} \stackrel{→}{OQ}=\stackrel{→}{OA}+\frac{1}{2}\stackrel{→}{AB} \\ =\stackrel{→}{OA}+\frac{1}{2}\left(\stackrel{→}{OB}-\stackrel{→}{OA}\right) \\ =\frac{1}{2}\left(\stackrel{→}{OA}+\stackrel{→}{OB}\right) \\ =\frac{1}{2}\left(\stackrel{→}{A}+\stackrel{→}{C}+m\stackrel{→}{A}\right) \end{gathered}$$

See that

$$\begin{gathered} \stackrel{→}{PQ}=\stackrel{→}{OQ}-\stackrel{→}{OP} \\ =\frac{1}{2}\left(1+m\right)\stackrel{→}{A} \end{gathered}$$

This implies $PQâˆ\yen OA$

Since $CBâˆ\yen OA$

This implies $PQâˆ\yen CB$

Hence, median is parallel to the two bases.

Now,

$$\begin{gathered} \left|\stackrel{→}{PQ}\right|=\frac{1}{2}\left(1+m\right)\left|\stackrel{→}{OA}\right| \\ =\frac{1}{2}\left(\left|\stackrel{→}{OA}\right|+m\left|\stackrel{→}{OA}\right|\right) \\ =\frac{1}{2}\left(\left|\stackrel{→}{OA}\right|+\left|\stackrel{→}{CB}\right|\right) \end{gathered}$$

Thus, the length of median is half the sum of length of the parallel sides.

Write PQ in vector form

$$\begin{gathered} \stackrel{→}{r}=\stackrel{→}{OP}+t\stackrel{→}{OA} \\ =\frac{1}{2}\stackrel{→}{C}+t\stackrel{→}{A} \end{gathered}$$

Let OB and PQ intersect at R and it diveides the line in the ratio $\mathrm{λ}:1$.

Thus, $\stackrel{→}{OR}=\frac{\mathrm{λ}\left(\stackrel{→}{OB}\right)}{1+\mathrm{λ}}$

This will satisfy the equation of PQ

So

$$\begin{gathered} \frac{\mathrm{λ}\left(\stackrel{→}{OB}\right)}{1+\mathrm{λ}}=\frac{1}{2}\stackrel{→}{C}+t\stackrel{→}{A} \\ \frac{\mathrm{λ}\left(\stackrel{→}{C}+m\stackrel{→}{A}\right)}{1+\mathrm{λ}}=\frac{1}{2}\stackrel{→}{C}+t\stackrel{→}{A} \end{gathered}$$

Thus, $\frac{\mathrm{λ}}{1+\mathrm{λ}}=\frac{1}{2}$

This implies $\mathrm{λ}=1$

Thus, R divides the line in the ration 1:1.

Hence, PQ bisects the line segment.