In this ‘proof’ the mistake is made when they have (a – t/2)^2 = (b – t/2)^2 and then take the liberty of ignoring the signs, and simply rooting.
For example, sure, it could be that in some cases a-(t/2) = b-(t/2) => a=b, which means that a-b=0, so we multiplied by 0 to begin with.
The other option is that a-(t/2)=-(b-(t/2)) => a-(t/2)=(t/2)-b => a+b=t, as we’d expect. (There are two more, but they’re just negative versions of the two cases that I stated, so the result is the same).
It’s clever, though.