how is it proved $\frac{Q_c}{Q_h}=\frac{T_c}{T_h}$? I know how to prove that the efficiency is $1-\frac{Q_c}{Q_h}$, but how do I go from that to $1-\frac{T_c}{T_h}?$ i.e. what is the proof that $\frac{Q_c}{Q_h}=\frac{T_c}{T_h}$?
 A: Consider a Carnot heat cycle as given below:

Now the Equation of efficiency, denoted by $\boldsymbol \eta$ for a Carnot cycle is given by:
$$\boldsymbol \eta = \frac{\boldsymbol W}{\boldsymbol Q_R} = \frac{\boldsymbol Q_H - \boldsymbol Q_c}{\boldsymbol Q_H} = \boldsymbol 1-\frac{\boldsymbol Q_C}{\boldsymbol Q_H}$$  

In an isothermic  process, 
$\Delta U = 0$;
$Q=W=nRT \log \frac{\boldsymbol V_2}{\boldsymbol V_1}$

For Isothemic process  $A \rightarrow  B$ and $C \rightarrow  D$
$${\boldsymbol Q_c=nRT_c \ln \frac{V_B}{V_A}}$$ 
$${\boldsymbol Q_H=nRT_H \ln \frac{V_D}{V_C}}= - {nRT_H \ln \frac{V_C}{V_D}}$$
Now by Dividing $\boldsymbol Q_c$ by  $ \boldsymbol Q_H$, we have
$$\frac{Q_c}{Q_H}= \frac{nRT_c \ln \frac{V_B}{V_A}}{-nRT_H \ln \frac{V_C}{V_D}}  = \frac{T_c \ln \frac{V_B}{V_A}}{-T_H \ln \frac{V_C}{V_D}} \tag{1}$$
Now from Adiabatic processes $B \rightarrow  C$ and $D \rightarrow  A$
$$T_CV_B^{\gamma-1} = T_HV_C^{\gamma-1} \tag{2.1}$$
$$T_HV_D^{\gamma-1} = T_CV_A^{\gamma-1} \tag{2.2}$$
Now by dividing $eq-(2.1)$ by $eq-(2.2)$, we have
$$\frac {T_CV_B^{\gamma-1}}{T_CV_A^{\gamma-1}} = \frac {T_HV_C^{\gamma-1}}{T_HV_D^{\gamma-1}}$$
$$\frac {V_B^{\gamma-1}}{V_A^{\gamma-1}} = \frac{V_C^{\gamma-1}}{V_D^{\gamma-1}}$$
Which means, $$\frac {V_B}{V_A} = \frac{V_C}{V_D} \tag{3}$$
Now the final part:
putting relation from $eq-(3)$ in $eq-(1)$, we have
$$\frac{Q_c}{Q_H}= - \frac {T_C}{T_H}$$
By taking the absolute values, 
$$\color{green} {\left| \frac {Q_c}{Q_H} \right| = \frac {T_C}{T_H}}$$
Now we can say that,
$$\boldsymbol \eta =  1-\frac{Q_C}{Q_H} = 1-\frac{T_C}{T_H}$$
