A question on voltage divider resistances used in BJT biasing
I don't understand why R2 and R1 parallel resistance should be less
  than or equal to one-tenth the dc?


The base always takes "some" current in order to produce an amplified current through the collector. Because the base takes current it in effect places a resistor in parallel with R2 and, the effect of this is seen to lower R2 a little bit. Therefore, if the base takes 0.01 mA then one rule of thumb suggests that the current flowing into R2 should be 0.1 mA.

This prevents the set-point voltage at the base being too much lower than the theoretical set-point determined soley by R1 and R2.

But of course, this is just a rule of thumb and some people use a different rule of thumb that says that "the parallel resistance of R1 and R2 should be 10x lower than the effective resistance looking into the base".

This slightly different rule of thumb has a slightly different result for the nominal values of R1 and R2 but, that doesn't matter because, there can be a wide range of R1 and R2 values that suit a particular amplifier circuit. It's not an exact piece of physics or science.

Given that the first rule of thumb says R1 and R2 should consume 0.1 mA and given the supply voltage is 10V, the series resistance of R1 + R2 should be 100k ohm i.e. 50k each (and not 100k as per your circuit). That of course is not a big-deal.

The 2nd rule of thumb would work like this; the base is at approximately 5V and consumes 0.01 mA therefore its resistance is effectively 500k ohms therefore, the combined parallel resistance of R1 and R2 should be no greater than 50kohm and of course R1 = R2 = 100k satisfies that but, given the "no greater than" clause, so does R1 = R2 = 50k.