Is similar triangles have equal areas a proposition? Suppose it is a proposition. So we have


*

*The conversion proposition is if two triangles have equal areas, then there are similar.

*The inversion proposition is that if two triangles are  not similar, then their areas are not equal.

*The contraposition is that if two triangles have different areas, then they are not similar.


As you know, they are both false. So two triangles  are similar is equivalent to they have same area.
We know this is not true.
So the original statement is not a proposition.
Another ques:
If $x>3$ then $x>5$.
Is this a proposition?
We know we can prove the equivalence of the contraposition and the original proposition by comparing the truth table. But suppose we are deal with the statement if P then Q where P and Q are not proposition. For instance, if $x>5$ then $ x>3$ is a true proposition while $x>5$ is not a proposition. 
 A: In logic, the Converse 

of a categorical or implicational statement is the result of reversing its two parts. For the implication $P → Q$, the converse is $Q → P$. For the categorical proposition All $S$ are $P$, the converse is All $P$ are $S$. In neither case does the converse necessarily follow from the original statement.

Thus if the original proposition is "similar triangles have equal areas", we can translate it as: 

$∀x∀y(\text {Tr}(x) ∧ \text {Tr}(y) \to (\text {Sim}(x,y) \to \text {SameArea}(x,y)))$

in which case we can "convert" it to : $∀x∀y(\text {Tr}(x) ∧ \text {Tr}(y) \to (\text {SameArea}(x,y) \to \text {Sim}(x,y))).$
Both are propositions in the "logical" sense, i.e. they have a definite truth value. 

"If $x > 3$, then $x > 5$" is not a proposition because it has no definite truth value: its truth value depends on the value assigned to $x$.
It is an open formula.
But if we quantify it, what we get is a sentence : $\forall x ((x > 3) \to (x > 5))$, which is FALSE in $\mathbb N$.
