How is definition of cutoff frequency consistent with KVL?
simulate this circuit &ndash; Schematic created using CircuitLab

Applying KVL, 

$$
V_{in} = V_{R} + V_{C} = I\cdot Z_C + I\cdot Z_R \\ 
\implies I = \frac{V_{in}}{Z_R+Z_C}
$$

Setting \$f=\frac{1}{2\pi RC}\$, we have 

$$
Z_C=\frac{1}{j2\pi fC}=\frac{1}{j2\pi \frac{1}{2\pi RC} C} = \frac{R}{j} = -jR
$$

In the low-pass filter, the output voltage is taken across the capacitor:

$$
V_{out,lpf}=I\cdot Z_C = V_{in} \cdot \frac{Z_C}{Z_R+Z_C} = V_{in} \cdot \frac{-jR}{R-jR} = V_{in} \cdot \frac{-j}{1-j} \\
\implies \left|\frac{V_{out,lpf}}{V_{in}}\right| = \left|\frac{-j}{1-j}\right| = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}}
$$

In the high-pass filter, the output voltage is taken across the resistor:

$$
V_{out,hpf}=I\cdot Z_R = V_{in} \cdot \frac{Z_R}{Z_R+Z_C} = V_{in} \cdot \frac{R}{R-jR} = V_{in} \cdot \frac{1}{1-j} \\
\implies \left|\frac{V_{out,hpf}}{V_{in}}\right| = \left|\frac{1}{1-j}\right| = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}}
$$