Help with a geometry question

[prisma jasper]

Hey, i need some help with a geometry question, i've been struggling with this for a decent bit of time now.
Prove that in any acute triangle, the circumscribed circle's center point to one of the vertexes is perpendicular to the chord of the other two vertexes height points
Here's a drawing i did of the problem aswell:

[pastel schooner]

@prisma jasper what did you try?

[prisma jasper]

nothing, that's what i'm stuck on, just did a drawing of the problem

[pastel schooner]

I mean i see a pretty trivial solution. You just have to find it

[prisma jasper]

i've asked some of my friends aswell, and none of them knew

if you could give me any leads, that would be really helpful

[pastel schooner]

Try a little angle chasing

A big lead ||prove angle OCA = 90 - angle CHI||

[prisma jasper]

alright, thanks, i'll let you know if i came up with something

[pastel schooner]

Im asuming you know cyclic quads?

@prisma jasper ?

[prisma jasper]

Yep, sorry was just a bit stuck on the english

I'm assuming you need to mirror H onto F?

cause that way you'd get a cyclic quad

Like this

@pastel schooner

[pastel schooner]

Oh no

Hint : ||ABIH is cyclic quad||

[prisma jasper]

Ohh

let me try

[prisma jasper]

Oh i get that, and the opposing angles add up to 180 degrees, and that ABI + AHI = 180, also that AHI = 180-CHI, so then ABI angle must be equal to CHI angle, how do i proceed though?

[prisma jasper]

Hi, I'm finished with the proof, there's just one question I have, How do I prove that ABIH is a cyclic quad? @pastel schooner