hard boi dometry

[quiet jackal]

triangle abc has orthocenter H and circumcenter O. If D is the foot of the perpendicular from A to Bc, then AH=8 and hD=3. If <AOH=90, find bc^2

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[serene seal]

So D is BOTH orthocenter and circumcenter?

[quiet jackal]

nah

D is the foot of perpendicular

[serene seal]

I meant O

[wide heart]

I assume "Hand" stands for "H and".

[quiet jackal]

no..

im pretty sure

H is orthocent

oh yeah

lemme ed it it

[serene seal]

Oh ok

I’ll try drawing it out when I get home

[quiet jackal]

any hints

on how to do it

or starts

[serene seal]

Maybe try splitting the triangles into a bunch of right triangles

You’ll definitely get similar triangles

[quiet jackal]

well

that kinda seems complicated

what i did so far

is extend AO

to create similar triangle

with AOH

so i extended to lets say X

AXD is similar to AOH

which also means <axd=90

but

how am i supposed to be bc^2

thats so

like

unreasonable

ohh

or maybe

i extend AO until it hits CD

[serene seal]

Find bc

[quiet jackal]

yeah

but still

thats unreasonable

cuz i dont know any angles

sides

i wonder if after i do that

if the new larger right triangle