Bonne Chance!

We can calculate the expected area for the two shapes. We want to aim at the one with the largest area to have the highest probability of hitting it. WLOG, let $r=1$ and let $O$ be the centre of the circle. We begin with the triangle :

Let $\angle CAB= \theta \Rightarrow \angle COB = 2 \theta$ because angle at the centre is twice angle at the circumference.

We calculate the area of $[\triangle ACB]= [\triangle COA]+[\triangle COB]=\frac{1}{2} \times 1 \times 1 \times \sin{(180^{\circ}-2\theta)}+\frac{1}{2} \times 1 \times 1 \times \sin{(2\theta)}=\sin{(2 \theta)}$

So the expected value of the area is $\frac{\int_{0}^{\frac{\pi}{2}}{\sin{(2 \theta)} d \theta}}{\frac{\pi}{2}-0}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}$ .

Now we move on to the rectangle :

If the height of the rectangle is $h$ we have the width of the triangle as $2 \sqrt{1-h^2}$ by Pythagoras' theorem on $\triangle FOM$ where $M$ is the midpoint of $FG$ .

The area of the rectangle therefore is $2h \sqrt{1-h^2}$ .

The expected value is: $\frac{\int_{0}^{r}{2h \sqrt{1-h^2} dh}}{1-0}=\frac{2}{3}$ .

As $\pi>3 \Rightarrow \frac{2}{\pi}<\frac{2}{3}$ so the expected area of the rectangle is bigger so we want to aim at the rectangle .

Area of triangle <= r^2 , Area of rectangle < 2r^2 , greater the area of a polygon greater the chances of dart hitting the polygon , so the answer is rectangle.