Bisecting Triangle

Let the length be $l$ and width be $w$ of the given right angled triangle.

Case 1 : When the length $l$ and width $w$ are equal then ratio is found to be $1:1$ reference angle $\theta$ is $45^\circ$ as $l$ = $w$ or, $tan\theta = \frac{w}{l}$ So $\theta = 45^\circ$

Case 2 : When the reference angle (\theta) is bisected into two equal angles then it( $\theta$ is splited into $22.5^\circ$ each.

So $tan22.5^\circ = \frac{sin22.5^\circ}{cos22.5^\circ}$

We have : $2sin^2\theta^\circ = 1- cos\theta$

$2cos^2\theta^\circ = 1+ cos\theta$

$tan22.5^\circ = \sqrt{\frac{1-cos45^\circ}{1+cos45^\circ}}$

On evaluting we get

$tan22.5^\circ = \sqrt{\frac{2-\sqrt2}{2+\sqrt2}} = \sqrt2 - 1$

Therefore, ratio of length and breadth $w'$ (new obtained width after bisecting the $\theta$ is

$\frac{l}{w'} = \frac{1}{tan22.5^\circ}$

$\frac{l}{w'} = \boxed{1: (\sqrt2 -1)}$

Alternatively :

Let bisected angles be $A$ and $B$ . Then

$tan(A+B) = tan\theta$

$tan(B + B) = 1$

$\frac{2tanB}{1-tan^2B} = 1$

$2tanB =1- tan^2B$

$tan^2B + 2tanB -1 = 0$

Solving for $B$ by using quadratic formula

$B = \frac{-2± \sqrt{4 -4(-1)}}{2}$

$B = \frac{2(1±\sqrt2)}{2}$

$B = -1±\sqrt2$

Since $B$ is an acute angle so $B = \sqrt2 -1$

so required ratio is found be $l : w' =\boxed{ 1 : (\sqrt2 -1)}$