Let $ABC$ be any triangle and let $O$ be a point on the line segment $BC$. Show that there exists a line parallel to $AO$ which divides the $\Delta ABC$ into two equal areas.
I have tried by rough geometric figure but failed. I got this from a well known book. Please help me to solve this.
In this solution, I will use area notation as []. In other words, I will express the area of a triangle $XYZ$ as $[XYZ]$.
Note that if $O$ is the midpoint of $BC$, we are done because $AO$ divides $\Delta ABC$ into two equal areas.
Now, suppose without loss of generality that $OB>OC$. We know that $[AOB]>[AOC]$. In addition, we know that $\frac{[AOB]}{[ABC]}=\frac{BO}{BC}$.
Now, consider points $M$ and $N$ on $BA$ and $BO$ respectively such that $MN$ is parallel to $AO$. We want to show that points $M$ and $N$ exist such that $\frac{[BMN]}{[AOB]}=\frac{BC}{2BO}$. Note that since $OB>OC$ and $OB+OC=BC$,$\frac{BC}{2BO}<1$.
However, we know that $\Delta MNB$ is similar to $\Delta AOB$ since $MN$ is parallel to $AO$. Therefore, $\frac{[BMN]}{[AOB]}=(\frac{MN}{AO})^2$. However, if we select $M$ and $N$ such that $\frac{BM}{BA}=\frac{BN}{BO}=\sqrt{\frac{BC}{2BO}}$ (which is possible since $\frac{BC}{2BO}<1$), we have $\frac{[BMN]}{[AOB]}=(\frac{MN}{AO})^2=\frac{BC}{2BO}$, so we have proved that $M$ and $N$ exist such that $\frac{[BMN]}{[AOB]}=\frac{BC}{2BO}$
Since $\frac{[AOB]}{[ABC]}=\frac{BO}{BC}$, we have $\frac{[BMN]}{[AOB]}\cdot\frac{[AOB]}{[ABC]}=\frac{BC}{2BO}\cdot\frac{BO}{BC}=\frac{1}{2}$, so we have proved that there exists points $M$ on $AB$ and $N$ on $BO$ such that $MN$ is parallel to $AO$ and $BMN$ has half the area of $ABC$.
That just follows from continuity. Let $x$ a point on the $BC$ line and $\ell_x$ a line parallel to $AO$ through $x$. Let $r_x$ be the ratio between the area of $ABC$ on the right of $\ell_x$ and the area of $ABC$. As $x$ travels on $BC$, $r_x$ smoothly goes from $0$ to $1$. It follows that at some point $r_x=\frac{1}{2}$.
As an alternative, assume that $\frac{[AOB]}{[ABC]}=r>\frac{1}{2}$. If you pick $O'\in BC$ and $A'\in AB$ such that $\frac{BA'}{BA}=\frac{BO'}{BO}=\frac{1}{\sqrt{2r}}$ then $O'A'\parallel OA$ by Thales' theorem and $[BA'O']=\frac{1}{2}[ABC]$ as wanted.
Here it is a straightedge and compass construction:
This is certainly true, just by continuity - we can move such a parallel line so that it divides the triangle in any proportions we chose.
If $O$ is equidistant from $B$ and $C$, the triangle area is already bisected.
Otherwise, assume by labelling that $O$ is closer to $C$, and define $r := \frac{\large OB}{\large BC}$. Define $D$ on $BC$ as where the parallel line must cross $BC$ to bisect the triangle area, and $E$ as the crossing point of that line on $AB$. We can find the location of $D$ as follows:
The parallel line closer than $AO$ to $B$ will create a triangle $\triangle DEB$ similar to $\triangle AOB$ that has reduced base and also reduced height in proportion. In fact $\text{ area } DEB = \left (\frac{\large DB}{\large OB}\right )^2\cdot \text{ area } AOB$ $ = \left (\frac{\large DB}{\large OB}\right )^2\cdot r \cdot\text{ area } ABC$. So we need $\left (\frac{\large DB}{\large OB}\right )^2\cdot r = \frac 12$, so place $D$ such that $DB^2 = \frac{\large OB^2}{\large 2r}$ $ = \frac 12OB.BC$
The existence follows from Jack's answer and here's a proof of how to explicitly constuct such a line. Let the line $l$ be parallel to $AO$ and let it divide the triangle in two equal parts in terms of area. WLOG let the line $l$ intersects $AC$ and $BC$ in $A_1$ and $O_1$. Then we have that: $$\frac 12 [ABC] = [CA_1O_1] \iff \frac{A_1C}{AC} = \frac{BC}{2CO_1}$$
Now from the parallelity we have; $$\frac{O_1C}{OC} = \frac{A_1C}{AC} = \frac{BC}{2CO_1} \implies CO_1^2 = OC \cdot CM$$
where $M$ is the midpoint of $BC$. To construct the length $CO_1$ just draw a circle with diameter $CO$ and draw a perpendicualr line to $BC$ at $M$. It will intersect the circle at two points $P,Q$. Now draw a circle with radius $CP$ and center $C$. The intersection of this circle and $BC$ is the point $O_1$