Here is a problem in which it's advantageous to use a Thevenin equivalent circuit. We want to determine the value of RL that will cause ib = 2 mA. (This is problem P5.5-12 from Introduction to Electric Circuits, 5e by R.C. Dorf and J.A. Svoboda.) We will solve this problem by finding a Thevenin or Norton equivalent circuit for the part of the circuit that is left of the terminals and then replacing that part of the circuit by its Thevenin or Norton equivalent circuit.
The row of buttons below the circuit correspond to various calculations that are used to find or to use a Thevenin or Norton equivalent circuit.
Thevenin and Norton equivalent circuits involve three parameters:
The buttons labeled "Voc", "Isc" and "Rth" show how to find a Thevenin or Norton equivalent circuit for the part of the circuit that is left of the terminals:

Analyze this circuit, e.g. by writing and solving a mesh equation, to verify the value of Rth.
As expected (Section 5.5 of Introduction to Electric Circuits by RC Dorf and JA Svoboda), Voc, Isc and Rth are related by
The buttons labeled "Thevenin" and "Norton" show the circuits that result when the part of the circuit that is left of the terminals is replaced by its Thevenin or Norton equivalent circuit.
Left-click on the "Thevenin" button to see the circuit that we get when we use the Thevenin equivalent circuit. Analysis of this circuit shows

Next, left-click on the "Norton" button to see the circuit that we get when we use the Norton equivalent circuit. Analysis of this circuit shows

Of course, we expect both of these calculations to yield the same value of RL.
Finally, left-click on the "circuit" button to return to the original circuit. Verify that setting RL = 0 does indeed cause ib=2 mA.