Art of Electronics zero-crossing circuit question
If we assumed that we have an Ideal comparator with an open collector output type. We can find the threshold voltages when we try to find the answer to this simple question.

At what voltage at the input, the voltage at noninverting input (\$V_P\$) equals the voltage at inverting input (\$V_N\$)
Hence when the output is at high state the voltage at the noninverting input is: 

\$ V_P = V_{CC}*\frac{R_6}{R_6+R_5+R_7} \approx +5\textrm{m}V   \$  

Therefore we need \$+5\textrm{mV}\$ at the inverting input (\$V_N\$) to trigger the circuit. 

Because of the resistors at the input, we do get \$+5\textrm{mV}\$ at the inverting input when the input voltage reaches 

\$Vth1 = \frac{(R1 + R2) R3 V_P + (R1 + R2 + R3) R4 V_P - (R1 + R2) R3 V_{CC})}{R3 R4}=29.66\textrm{mV}\$

So, now our comparator output is at the low state hence the voltage at noninverting input is \$V_P = 0V\$. And again we need to find for what voltage at the input terminal the inverting input being equal to that at the noninverting input (\$V_P = V_N = 0V\$). 

\$Vth2 = - \frac{R_1+R_2}{R_4}* V_{CC} = -21.85\textrm{mV}\$ 

As you can see the hysteresis is not fully symmetrical.