Let$\ P_0 $be our original proposition and$\ P_1 :a<b $ and$\ P_2:a<c $ the statements that are equivalent to$\ P_0 $. Now, if it is known that$\ b<c $, then there is an interval of values of$\ x $ such that$\ b<x<c $. Since both$\ P_1 $ and$\ P_2 $ are necessary and sufficient for the truth of$\ P_0 $, doesn't it mean that for the latter to be true it is only needed that$\ a $ assumes any value respectively$\ <b $ and$\ <c $, and we can't just say that$\ a$ must be$\ <b $? If so, recall $\ b<x<c $. If$\ a $ is in this interval, then$\ P_1 $ is false and $\ P_2 $ is true, leading to the contradictory result that $\ P_0 $ is at the same time true and false. Thus, may we conclude that$\ P_0 $ is false?
I hope I made myself clear.
EDIT: I'm putting here some more details from my answer to Nick and paw88789. what I'm saying is that, known that $\ P_2 $ is a necessary and sufficient condition for $\ P_0 $, one should be able to pick any $\ x<c $ for $\ P_0 $ to be true. But, as it turns out, it's not like that. In fact, if $\ b<a<c $ we have that $\ P_0 $ is both true and false, which is a contradiction, so they're all false. Alternatively, we may note that it must be $\ a<b $ to avoid the "contradictory interval". Thus, $\ P_2 $ loses its sufficiency, which is absurd since it was demonstrated. This is why I thought of concluding that all the propositions are false.
My latest guess is that since both $\ P_1 $ and $\ P_2 $ have been proven to be equivalent to$\ P_0 $, the truth of either of the three statements implies that of the others, and that's exactly because $\ a $ can't lie in the "contradictory interval", so showing that$\ a <c $ gives us that$\ a <b $ as well. That is, we can't deduce the falsity of the three statements.
You have a proposition $P_0$ and you say it is equivalent to $P_1:a<b$ and it is also equivalent to $P_2: a<c$. You say you also know $b<c$ and I assume they are all real numbers. This is not enough to conclude that these statements cannot all be true together, e.g. $(a,b,c)=(1,2,3)$ satisfies these conditions. It's different if $a$ is a free variable:
Let's look at $$S_1=\{a|a<b\}, S_2=\{a|a<c\}.$$ We know $S_1=S_2$ because $P_1$ is equivalent to $P_2$. We also know $\frac{b+c}{2}\in S_2\setminus S_1$ because $b<\frac{b+c}{2}<c$. We therefore have arrived at a contradiction. If you were attempting to do a proof by contradiction, assuming $P_1$ or $P_2$, knowing that $b<c$ and that $P_1$ is equivalent to $P_2$, then yes you can conclude that all of $P_0, P_1, P_2$ are false. Otherwise, you have derived a contradiction, hopefully a mistake or misunderstanding somewhere, otherwise we are all in trouble...