NodeMCU GPIO pin Still Has Power?

Question

So I have a NodeMCU board and I am trying to create a garage door opener. Which I was partially successful with. To explain, I was able to control the garage door using the NodeMCU board and open and close it successfully. However, the issue I am running into is the second I connect the garage door openers pins into my breadboard I cannot open and close the garage with the regular garage door opener button on the wall. From what I understand, it seems as the GPIO pin even in the LOW state (which the led is turned off) has power. My question is, is there any way to completely turn off the GPIO without having to use a relay? Thank you to everyone in advance!

This is the code I am using to turn on and off the GPIO feel free to let me know if I am doing something incorrectly.

digitalWrite(outputD5, LOW); 
digitalWrite(outputD5, HIGH); 
delay(1000); 
digitalWrite(outputD5, LOW); 

Answer 1

The solution for this question would be to switch both pins to input that way no signal is sent out, as an added safety measure set both pins to low. When you want to toggle them change the pins to output and high and then revert them to the original after a delay or what not.

Answer 2

The answer will be No !

If the pin has been configured as an OUTPUT with pinMode(), its voltage will be set to the corresponding value: 5V (or 3.3V on 3.3V boards) for HIGH, 0V (ground) for LOW.

Make sure you are using appropriate resistor in your schematics and try probing the output pin.