If you make a straight cut through a square, one part can always be made to cover the other. (This is true by symmetry if the cut goes through the centre, and if it doesn't, you can shift it to the centre while taking from one part and giving to the other.)
However, if you cut an equilateral triangle, it may or may not be the case that one part can be made to cover the other. In some cases it may depend on whether we're allowed to flip the parts; I'll leave that to you in case one or the other version has a more elegant solution.
- How can the cuts that allow one part to cover the other best be characterized?
- What is the probability that a random cut will allow one part to cover the other?
Of course we need to specify a distribution for the cuts, and again I'll leave you to choose between two plausible distributions in case one yields a nicer result: Either Jaynes' solution to the Bertrand "paradox" (i.e. random straws thrown from afar, with uniformly distributed directions and uniformly distributed coordinates perpendicular to their direction), or a cut defined by two independently uniformly distributed points on two different sides of the triangle.
Update: I've posted the case without flipping as a separate question.
If we allow flipping, an answer to the first question will be (as suggested by Thomas Ahle in the comments):
The "only if" part
This can be shown by contraposition. So let us assume that not all three altitudes are cut within the triangle and prove that one part covers the other:
If (at minimum) one of the altitudes is left uncut within the triangle, this uncut altitude provides an axis of symmetry over which you can reflect one part of the triangle to cover the other part.
The "if" part
First note that if a cut passes through either a vertex and/or a midpoint (at least) one of the altitudes is not being cut within the triangle.
So the cut must pass between vertices and midpoints in order to cut all three altitudes within the triangle. Thinking a bit more about this we realize that then the situation can be rotated to fit the following diagram of the situation:
Where the cut has to enter through the red segment at one side and exit through the other red segment at the other side. This means that some kind of cut lying entirely below the blue equilateral at the top of the diagram has to be made.
But then it is clear on one hand that the bottom part cannot in any possible way cover even just the blue triangle, less so the top part of the cut. And the opposite is clearly also not possible since the bottom part has a full side length of the original triangle which is a distance that is nowhere to find in the top part of the cut.
The last part could have been put in more specific and technical terms, but I think that would blur the picture. In case someone disagrees, please suggest improvements or ask for clarification.
Probability figure
Let us place an equilateral triangle of side length $1$ (WLOG) inside a circle of diameter $2\cdot\sqrt 3/3$:
Then using the distribution given as method 2 in the OP slightly re-phrased, a chord can be chosen by rotating the circle by a uniformly randomly chosen angle and the choosing a point $E$ on the vertical diameter (the orange diameter in the diagram above) uniformly at random and drawing a horizontal line through that point.
Any line in the plane will form an angle $v$ within the interval $[0,\pi/6]$ with one of the altitudes of the randomly tilted equilateral triangle. WLOG assume the altitude in question is $BD$.
Now letting the point $E$ traverse the vertical diameter of length $2\cdot\sqrt 3/3$ the horizontal line will cut the triangle iff it cuts the vertical segment $AG$ which will be denoted by $w$ for future reference. So the probability that a horizontal line that cuts the circle (the event $\Omega$) also cuts the equilateral triangle (the event $\Delta$) for the given tilt angle $v$ will be: $$ P(\Delta\mid\Omega,v)=\frac{w}{2\cdot\sqrt 3/3}=\frac{\cos(v)}{2\cdot\sqrt 3/3} $$ where we have used that the side length of the equilateral was $1$ in order to establish that $w=\cos(v)$.
Next let us consider the probability that the horisontal line intersects all three altitudes (the event $\star$) given that it intersects the triangle. This can be expressed as the probability that it intersects the segment $z=DF$ given that it intersects the side $AC=1$. A simple use of trigonometry shows that: $$ P(\star\mid\Delta,v)=\frac z1=\frac{\sqrt 3}2\cdot\tan(v) $$
Finally, some tedious integration leads to the general statement integrating out $v$ providing the probability figure: $$ \begin{align} P(\star\mid\Delta)&=\dfrac{\int_0^{\pi/6}P(\star\mid\Delta,v)\cdot P(\Delta\mid\Omega,v)\ dv}{\int_0^{\pi/6}P(\Delta\mid\Omega,v)\ dv}\\ &=\dfrac{\frac 34-\frac 38\cdot\sqrt 3}{\sqrt 3/4}\\ &=\sqrt 3-1.5\\ &\approx 0.2320508 \end{align} $$
So this determines the probability that any one part will fail to cover the other part, and the corresponding probability that one part will cover the other under this distribution (event $\chi$) is: $$ P(\chi)=1-P(\star\mid\Delta)=2.5-\sqrt 3\approx 0.76794919 $$