In an isosceles right angled triangle, a straight line is drawn from the mid point of one of the equal sides to the opposite vertex. Then a pair of possible values of the cotangents of the two angles so formed at that vertex are

$\text{ Let }ABC\text{ be the triangle, right angled at }C\text{, and }D\text{ be the mid-point of }AC\text{. Join }DB\text{. }$$\text{ Since }AC=BC=x\text{, we have }$

$DC=\frac{1}{2}AC=\frac{1}{2}BC=\frac{x}{2}$

$\text{ Also }\mathrm{\angle }CAB=\mathrm{\angle }CBA={45}^{\circ }$

$\text{ If }\mathrm{\angle }DBC=\theta \text{ and }\mathrm{\angle }DBA=\varphi \text{, }$

$\tan \varphi =\tan \left({45}^{\circ }-\theta \right)=\frac{1-\tan \theta }{1+\tan \theta }$

$\tan \varphi =\frac{DC}{BC}=\frac{x/2}{x}=\frac{1}{2}$

$\text{ or }\tan \varphi =\frac{1-1/2}{1+1/2}=\frac{1}{3}$

$\therefore \cot \theta =2,\cot \varphi =3$