Power supplied by a Thevenin source
$$P_{Thevenin} = V_{Th}^2 \frac{1}{R_{Th}+R_L}$$
Power supplied by its equivalent Norton source
$$\begin{eqnarray} P_{Norton} & = & I_N^2 \frac{R_{Th} R_L}{R_{Th}+R_L} \\ & = & \left(\frac{V_{Th}}{R_{Th}}\right)^2 \frac{R_{Th} R_L}{R_{Th}+R_L} \\ & = & V_{Th}^2 \frac{R_L/R_{Th}}{R_{Th}+R_L} \end{eqnarray}$$
So power supplied by a Thevenin Source and its equivalent Norton source are equal only when \$R_{Th} = R_L\$.
Is it true or am I missing something?
Yes, you're missing something.
The Thévenin source includes both the voltage source AND the resistor, and the Norton source includes both the current source and the resistor.
In both cases, the power consumed by the internal resistor is not considered to be power "delivered" by the source. Instead, consider only the source's terminal voltage (where it connects to the load) and the current passing from the source to the load.
You'll find that they are exactly equivalent, regardless of the load. This is the whole point of having "equivalent" sources in the first place. You can replace any real source with its Thévenin or Norton equivalent and get identical system behavior. If each replacement source is exactly equivalent to the original, it follows that they must also be equivalent to each other.
