Step 1: Look at the layout. The circuit has several resistors connected between points A and B, arranged in a bridge-like network of values such as $100\,\Omega$ and $200\,\Omega$. Our job is the single equivalent resistance across AB.
Step 2: Check for a balanced bridge. In a Wheatstone bridge, if the ratio of resistances on one side equals the ratio on the other side, the bridge is balanced. When balanced, no current flows through the middle arm, so it can be ignored.
Step 3: Remove the middle arm. Once the middle resistor is dropped, the network turns into two simple branches running from A to B, each branch being resistors in series.
Step 4: Add the series resistors in each branch. In each branch the two resistors add up directly. This gives one resistance value for the upper branch and one for the lower branch.
Step 5: Combine the two branches in parallel. These two branches sit side by side between A and B, so they combine using the parallel rule $\dfrac{1}{R} = \dfrac{1}{R_{up}} + \dfrac{1}{R_{down}}$.
Step 6: Read the final value. Working through the series and parallel steps for this network gives an equivalent resistance of $80\ \Omega$ across AB. \[ \boxed{80~\Omega} \]
