Feeding a PCB dipole with single ended RF signal
First, make a stubline/series trans line matching circuit to match your 50Ω to 71Ω:



The lengths you need result from reading the ratio of \$\lambda\$ from a smith chart.

After that, you could implement a Rat-Race Coupler in PCB technology, matched to your 71Ω source:



In the naming above, you'd feed in your unipolar signal at P1, and attach your dipole to P2 and P4. (omit P3, or terminate it to ground with 71Ω – if your dipole is well-matched, the power leaving P3 should be 0.)

To give you a feeling of the sizes involved:

a \$\frac\lambda4\$ piece of microstrip line is 45mm long on 1.6mm thick FR4 (assuming an Er=4.7 and 35µm copper), so you your rat race ring would need to have a circumference of ca 275mm, corresponding to a diameter of 87.5 mm. That's not really small – but if you have PCB to spare (or can elegantly route this around your whole circuit!), that's a nice way of going at this.

If you need to save space, you need to be smarter about the rat race: You might want to replace the transmission line \$\frac34\lambda\$ element with an actual RC (lowpass!) that is equivalent in phase shift, for example.

Another option would be to go for a Wilkinson power divider (which is only \$\frac13\$ of the size), followed by a \$\frac\lambda2\$ phase shifting network on one side only.