Due DigitalRead() threshold voltage

I see that 5v platforms use a 3v digital threshold.

Does anyone know what the digital input threshold voltage is on Due (3.3v system)? There's probably some hysteresis too?

i would find it experimentally but I don't have the bits to hand.
Thanks,
Mark

V IL Input Low-level Voltage PIOA/B/C/D/E/F[0-31] -0.3 0.3 x V VDDIOV IH
Input High-level Voltage PIOA/B/C/D/E/F[0-31] 0.7 x V VDDIO V VDDIO+0.3V
Hysteresis Voltage V PIOA/B/C/D/E/F[0-31]
except PA0, PA9, PA26, PA29, PA30, PA31, PB14, PB22, PC[2-9], PC[15-24], PD[10-30],
PE[0-4], PE15, PE17, PE19, PE21, PE23, PE25, PE29 150 500 mV

Section 46.2

It should be pointed out that the ATmega family of processors happen to have guaranteed
thresholds of 1.5V and 3V when run at 5V, but in general 5V logic systems do not always recognise 3.3V as high. You go to the datasheet for this information - see Static discipline - Wikipedia

Just because this is a related question to do with the SAM3X and its voltages, if I am reading this right, the I/O pins according to 46.1/46.2 can handle up to 4v, 3.6v, or VVDDIO +0.3V (these were the main values that sounded like they might be right) before risking or causing damage? 3.6v sounds about right as it would provide about a 10% tolerance which if I'm not mistaken is pretty standard. Also would this be applied to the ground pin as well, or does the ground pin bypass the processor so that it would be able to take the full brunt of the 5v?
I am not used to reading these datasheets and am still familiarizing myself with the terminology, so I'm sorry if this seems like a no brainer. I just need to make sure before hooking up a sensor that runs 5v with a voltage divider dropping the max output of 5V to 3.333... before it reaches the I/O pin, and I have no idea what kind of threshold/tolerance I have with the I/O pins before I cause damage.
Thank you for your time

The I/O pins are limited to the supply range +/- 0.3V, ie it depends on the supply voltage.

Bringing any pin outside that range will start to cause the protection diodes to start
conducting (these are there to help protect against static electricity). Protection diodes
are small and cannot tolerate much current continuously, think < 1mA.

In general if there is any risk of an external signal exceeding the voltage range you need
protection circuitry, such as schottky diodes to back up the protection diodes.

For instance a large capacitor on an I/O pin will risk back-powering the whole device
on turn-off (sometimes analog pins have 100nF on them, this is safe, but large values
in the uF range are not a good idea).