Understanding Touch Switch with BJTs and Relay

We worked on this circuit today in one of my labs and it had my groupmates and I puzzled. We had a 1k resistor in series with an LED on the NO side of the relay.

I searched for the circuit online and found this website explaining, but his explanation does not make any sense to me. He says that when you activate T1 and then T4, this "energizes" the relay, thus activating the other circuit. This part makes sense, but the following has me confused. When you let off of the touch switch for T1 and T4, the light continues to stay on. By my understanding of relays and transistors, this doesn't make sense. The transistor is acting as a switch, providing ample current to the relay, but should it not then cut off when let off of the switch?

Secondly, T2 and T3 are activated to turn the relay off. This part makes a but more sense as the high current from the darlington transistor would draw all of the current from the relay, I believe.

• Just as a side note to the answer you already have, when the relay engages, the armature is pulled down towards the solenoid and shortens the magnetic path length of the solenoid and making it easier to "hold in." (This means, "less current is required" to keep it there.) So when T1 and T4 are no longer activated, R2 and D2's somewhat lower current (with the internal relay resistance in series, of course) is still sufficient to hold the relay. Darlington T2+T3 then can source current into R2, forcing its voltage drop to increase enough to leave still less current so the relay can release. – jonk Feb 9 at 7:39

This is a bit of a creepy circuit. It appears to work on the relay hysteresis. So the relay coil, in series with a diode D2 and the 100$$\\Omega\$$ resistor, receives enough current to hold the relay in, but not to actuate it.
When TP2 is bridged, the 100$$\\Omega\$$ resistor receives current from the Darlington pair T2/T3 and the relay current drops to a very low level, so it drops out.