In triangle $ABC$, $\space BE \space$ is a median, and $O$ the mid-point of $BE$. The line joining $A$ and $O$ meets $BC$ at $D$. Find the ratio $\space AO : OD \space$ (Hint: Draw a line through $\space E \space$ parallel to $\space AD$.)
using BPT in first triangle BEP, we get relation b/w OD and EP. Similarly using BPT in second triangle we get relation between AD and EP. Thus, ultimately we get relation between AD and OD. Using this we can find the ratio of AO and OD. Am I right?
Hints: $AE=EC \space, BO = OE, \space AD \mid\mid EF, \space OD \mid \mid EF$;
So, in triangle $\space ADC \space$ we have: $\space AD \mid \mid EF, \space$ hence by basic proportionality theorem we also have:
$$\frac{CE}{EA} = \frac{CF}{FD}.$$
Since $AE=EC, \space$ we will have:
$$1=\frac{CF}{FD} \iff CF=FD. \space \space \space \space \space \space \space (1)$$
Now, in triangle $\space BEF \space$ we have: $\space EF \mid \mid OD, \space$ hence by basic proportionality theorem we also have:
$$\frac{FD}{DB} = \frac{EO}{OB}.$$
Since $BO = OE, \space$ we will have:
$$\frac{FD}{DB}=1 \iff FD=DB. \space \space \space \space \space \space \space (2)$$
So, we know: $\space \boxed{CF=FD=DB \iff CF=FD=DB=\frac{1}{3}CB, \space CD=\frac{2}{3}CB.\space}$
Finally, in triangles $\space ADC \space$ and $\space BEF \space$ we also have:
$AD:CD=EF:FC \\ OB:OD=EB:EF$
$$ \iff EF=\frac{AD}{CD}FC, \space \space \space \frac{OB}{OD}=\frac{EB}{\frac{AD}{CD}FC}=\frac{EB\cdot CD}{AD \cdot FC}.$$
Since we have $\space OB=\frac{t_B}{2}, \space OD=y, \space CD=\frac{2}{3}CB, \space AD=x+y, \space FC=\frac{1}{3}CB, \space$ we can write:
$$\frac{\frac{t_B}{2}}{y}=\frac{t_B\cdot \frac{2}{3}CB}{(x+y)\cdot \frac{1}{3}CB} \iff \frac{1}{2y}=\frac{2}{x+y} \iff 4y = x+y\iff 3y =x \iff \boxed{\space \frac{x}{y}=3. \space}$$
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