Resistor simplification
The equivalent resistance (aka impedance) that can replace resistors in parallel can be written like this:

\$R_{parallel} = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+..}Ω\$

This means that if \$R_1 = 53Ω\$ and \$R_2= 205Ω\$ then you can replace both of those two with one resistor that has the value \$\frac{1}{\frac{1}{53}+\frac{1}{205}}=42.1Ω\$

Or it can be written like this if it's just two resistors in parallel. 

\$R_{parallel} = \frac{R_1×R_2}{R_1+R_2}\$

With the same example numbers as above, \$\frac{53×205}{53+205}=42.1Ω\$

For 3 resistors in parallel it's not \$\frac{R_1×R_2×R_3}{R_1+R_2+R_3}\$, it's \$\frac{R_1×R_2×R_3}{R_1×R_2+R_2×R_3+R_1×R_3}\$. 
I derived this expression from the first equation above (\$\frac{1}{...}\$).

The equivalent resistance that can replace resistors in series can be written like this:

\$R_{series} = R_1+R_2+...\$

When I write "..." I mean that the terms can be repeated if you have more resistors.



With the information above together with the schematic we can arrive to the conclusion:

The resistance between a and rotated b is equal to \$\frac{R_1×R_2}{R_1+R_2}\$

The resistance between rotated b and c is equal to \$\frac{R_3×R_4}{R_3+R_4}\$

The resistance between a and c is equal to \$\frac{R_1×R_2}{R_1+R_2}\$+\$\frac{R_3×R_4}{R_3+R_4}\$