Question

In an isosceles right angled triangle a straight line is drawn from the middle point of one of the equal sides to the opposite angle. Show that it divides the angle into two parts whose cotangents are 2 and 3.

Answer 1

Let ABC be the triangle, right angled at C, and D be the mid-point of AC. Join DB.

Since AC =BC =x, we have

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

Also $\angle CAB=\angle CBA={45}^0$

If $\angle DBC=\theta$ and $\angle DBA=\varphi$,

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

$tan\varphi =tan({45}^0-\theta )=\frac{1-tan\theta }{1+\tan \theta }$

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

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