CE : CO : CB = 1 : 2 : 4 (do you know why?)

EO : EB = 1 : 3 as a result

Since CH : CF = CO : CB = 1 : 2, then HO : FB = 1 : 2; this is due to midpoint theorem.

Because of this proof, we can conclude that triangles CHO and CFB are similar, with lines HO and FB being parallel as well.

Let's consider triangles EXO and EFB, knowing that lines XO and FB are parallel. Are you able to prove their similarity?

Since triangles EXO and EFB are similar,
XO/FB = EO/EB = 1/3

Since HO : FB = 1 : 2 and XO : FB = 1 : 3, then it follows that HO : XO : FB = 3 : 2 : 6, so...

...OX : OH = 2 : 3

Then it's clear that OX : XH = 2 : (3 - 2), or 2 : 1.

Thank you. Not sure why I couldn't establish that HO // FB. I'm embarrassed to be a H2 Further Math student when I can't even help my younger niece with this... -_-"

It’s not that immediately obvious because the proof for the midpoint theorem actually relies on the similarity of triangle proof which is least commonly used of the three approaches - the SAS version of similarity is not that commonly seen even in the O Levels.