$${If}\:{C}'\:{and}\:{B}'\:{are}\:{midpints}\:{of}\:{AC}\:{and} \\$$$${AB},\:{find}\:{cot}\left(\theta\right)\:{as}\:{a}\:{function} \\$$$${of}\:{angles}\:{B}\:{and}\:{C}. \\$$
$${answer}:\:{cot}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({cot}\left({B}\right)+{cot}\left({C}\right)\right) \\$$$${How}\:{can}\:{I}\:{get}\:{this}\:{result}? \\$$
$$\mathrm{cot}\:\theta=\frac{{PB}'}{{PM}} \\$$$$=\frac{\frac{\mathrm{1}}{\mathrm{2}}{B}'{C}'}{{PM}} \\$$$$=\frac{\frac{\mathrm{1}}{\mathrm{2}}\left({BC}−{B}'{C}'\right)}{{PM}} \\$$$$=\frac{\frac{\mathrm{1}}{\mathrm{2}}\left({BM}+{MC}−{C}'{P}−{PB}'\right)}{{PM}} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\frac{{BM}−{C}'{P}}{{PM}}\right)+\left(\frac{{MC}−{PB}'}{{PM}}\right)\right\} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cot}\:{B}+\mathrm{cot}\:{C}\right) \\$$$$\boldsymbol{{Plese}}\:\boldsymbol{{treat}}\:\boldsymbol{{it}}\:\boldsymbol{{as}}\:\boldsymbol{{a}}\:\boldsymbol{{hint}}\:\boldsymbol{{not}}\:\boldsymbol{{complete}} \\$$$$\boldsymbol{{solution}}. \\$$