Which is the most acidic proton of (3Z)-3-ethylidenecyclopent-1-ene-1-carbaldehyde?
If you draw curly arrows for the deprotonations, it becomes clear that only the pathway in green is productive due to the ability to form an extended enolate. 

If you deprotonated the green protons, the resulting negative charge is able to be delocalised through the alkene and onto the carbonyl, the product of which is essentially an extended enolate. Due to this stabilisation, the pka of the green protons is relatively low (the resulting conjugate base is stabilised). 

Deprotonation of the red proton however doesnt yield a charge which may be delocalised onto the carbonyl ( you would end up with a 5 valent carbon if you draw the arrows ). What you essentially end up with in this case then is an allyl anion (which itself is conjugated to the exocyclic alkene), which although stabilised by resonance isn't quite as favourable as in the green case (the pka of the red protons is higher).