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

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.

• In the solution, there was a mistake (it doesn't affect this part of the proof), which I have fixed, so this part also MAY be incorrect (it might be correct, but I just wanted to point out the possibility). If someone does prove/disprove that GB=HC=1, please post your steps. – Chirag Maheshwari Jan 6 at 4:20

$$\triangle ABD :$$ 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.