There are two steps to finding the Thevenin equivalent circuit: finding the Thevenin voltage and the Thevenin resistance.
Thevenin voltage is the voltage across the two points you interested in (Vin). In this case it is easy to calculate as there is no current flowing in the 43 and 60 \$\Omega\$ resistors thus no voltage drop. Thus the voltage at Vin is the same as the voltage form the source, 72 V.
Thevenin resistance is calculated by 'turning off' all independent current and independent voltage sources and calculating the resistance between the two points. Turning off a voltage source sets the voltage across it to 0, which results in a short (0 \$\Omega\$) in parallel with the 275 \$\Omega\$ resistor. Any resistor combined in parallel with a short results in a short, leaving you with the 43 and 60 \$\Omega\$ resistors now in series, giving a Thevenin resistance of 103 ohms.
Putting the two together gives you a voltage source of 72 V in series with a 103 \$\Omega\$ resistor for you Thevenin equivalent circuit.
That's simple: in high school they did not tell you the bad part of the story: Y-\$\Delta\$ transform.
The series parallel technique works well... Almost in all cases, but when you need to solve such a bad guy then you need to use all your weapons. You should read the wikipedia article, anyway given a three impedances circuit, either arranged as a star, i.e. a central node common only to the impedances and three external nodes, or as a triangle, i.e. three nodes where only two of the tree impedances connects, it is possible to transform it in the other fashion, that is transform a triangle in a star and vice versa. The formulas are all written on wikipedia, I am not copying them here, the interesting part is that with series, parallel and Y-\$\Delta\$ transforms you can solve almost any network.
In your particular case you'd trasform the ABD triangle in a star with tips in ABD and center c (that's a new node), then compute the series cA->AC and cD->DC and then the series Bc->cC. If you write it down you'll grasp this in seconds.
There also is another transform called star-mesh transform, which is a generalization of the Y-\$\Delta\$ transform. Please note that with more than three impedances, that is our case of interest, the transform is not reversible since it results in an increased number of impedances.
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I find for some students, the angled components are visually confusing. Try redrawing the same schematic with only horizontal and vertical components to see if this helps you to analyze it.
For example, if you start on the right and straighten out the 6k resistor, you will notice it is in parallel with the series combination of 1k and 2k. So you have 3k in parallel with 6k for a total of 2k. Now move to the left and you see that this 2k is in series with the 10k. Keep up this procedure to reduce the entire lattice to a single value.