Angles in gray triangles are $\alpha, \beta$ and $90^\circ$.
Catheti (green and red) are equal, when $\alpha=\beta$.

So, central rectangle can be square when $\alpha=\beta ~~$ too.

This problem is a particular case of a family of problems with broadly the same solution, so I will post this more general solution and then discuss particular instances of it.

Problem. $\angle BAC=3\angle CAD$; $\angle CBD=30^\circ$; $AB=AD$. What is $\angle DCA$?

Solution. Let $\alpha=\angle CAD$. $\triangle BDA$ is isosceles on base $BD$. Therefore $\angle DBA=\angle ADB=90^\circ-2\alpha$ and $\angle CBA=120^\circ-2\alpha$.

Let $E$ be on $BC$ such that $AE=AB$. Then $\triangle BEA$ is isosceles on base $BE$. Therefore $\angle AEB=\angle EBA=120^\circ-2\alpha$, so $\angle BAE=4\alpha-60^\circ$, so $\angle EAD=60^\circ$.

Therefore $\triangle AED$ is equilateral, so $\angle EAC=60^\circ-\alpha=\angle ACE$, so $\triangle CAE$ is isosceles on base $CA$, i.e. $CE=AE=DE$, so $\triangle CDE$ is isosceles on base $CD$. $\angle CED=2\alpha$, so $\angle DCE=90^\circ-\alpha$, so $\angle DCA=30^\circ$, which solves the problem. Note that $\angle DCA$ is independent of $\alpha$.

To adapt this to the current problem, relabel from $ABCD$ to $BCDA$ and specify $\alpha=19^\circ$.

If $\alpha$ is specified as $20^\circ$, and $\angle DBA$ as $50^\circ$, then the problem is [Langley]. $AB=AD$ is easily seen, and the proof proceeds as above. The above proof, but with angles as specified in Langley's problem, is due to J. W. Mercer.

If $\alpha$ is specified as $16^\circ$, then the problem is that at gogeometry. The point-lettering is the same, but the diagram is flipped.

[Langley] Langley, E. M. "Problem 644." Mathematical Gazette, 11: 173, 1922, according to David Darling

## Best Answer

$AH \perp AR$

$AB=AD$ (Sides of Square)

$\angle ADH =\angle DAR$(alt. int $\angle 's, DH \parallel AR$)

$\angle ABR + \angle BAR = \angle DAR + \angle BAR=90^o$

Hence,

$\angle ABR = \angle DAR $ and $\angle ABR = \angle DAR $ and

$\angle ADH =\angle ABR$

and

$\angle ADH + \angle DAH = \angle ABR + \angle BAR=90^o$

Hence, $\angle DAH =\angle BAR$

$\triangle ADH \cong \triangle ABR$

Therefore $AH=AR$

Two consecutive sides both perpendicular and equal.

$AHKR$ is a square