Calculate Thevenin voltage in a Common Emitter Circuit
It's been +25 years but I gonna give it a try ...

\$R_{th}=\dfrac{V_{oc}}{I_{sc}}\$



Calculating \$I_{sc}\$ :

\$I_{b}=\dfrac{10.7V - 0.7V}{10k\Omega}=1mA\$

\$I_{c}=I_{b}\cdot\beta =1mA\cdot100=100mA\$

\$I_{sc}=I_{b}+I_{c}=1mA+100mA=101mA\$



Calculating \$V_{oc}\$ :

\$-10.7+10000\cdot I_{b}+0.7+1000\cdot(\beta+1)\cdot I_{b}=0\$

\$10000\cdot I_{b}+1000\cdot101\cdot I_{b}=10\$

\$I_{b}=\dfrac{1}{11100} A\$

\$I_{e}=(\beta+1)\cdot I_{b}=\dfrac{101}{11100}A\$

\$V_{oc}=I_{e}\cdot 1k\Omega=\dfrac{101\cdot1000}{11100}V=\dfrac{1010}{111}V\$



\$R_{th}=\dfrac{V_{oc}}{I_{sc}}=\dfrac{\dfrac{1010}{111}V}{101mA}=90.09\Omega\$



PS : It's 5:49AM here, time to get some sleep.