$2020$ AIME I Problem $13$ Identifying Cyclic Quadrilaterals

Question

Link: https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_13#Solution_1

I will be referring to Solution 1.

I understand the solution completely, except for the part where it says

$GB=HC=1$

I don't understand how this is true. I tried to prove it using Power of a Point, angle chasing, and Law of Sines/Cosines, but none of these steps got me far.

Can someone please explain why it is true?

Thank you.

Answer 1

$\triangle ABD : $ the angle bisector of $\angle B$ meets perpendicular bisector of opposite side $AD$, on its circumcircle. This point is $E$. So $ABDE$ is cyclic as is $ACDF$.

$$\angle AHB = \dfrac{1}{2}\angle AED = 90-\dfrac{B}{2}=\angle HAB$$

Conclude $AB=BH$ and similarly $CG=AC$ from which $GB=HC=1$ follows.