@HrishabhNayal 74 8 74

@RajdeepSindhu 53 19 , right?

@PrateekMourya wai... what?

@HrishabhNayal Konami code

Like*

How to solve this >_<

@Wolgwang Did you know about this prior to his message?

@RajdeepSindhu Obviously -_-

I had tried this on stackexchange hat page.

@Wolgwang (I'm using the alphabets from A to D clockwise, not the diagram you posted) Draw perpendiculars AM, CN onto BD. $\angle AOM = \angle CON$ and $\angle AMO = \angle CNO = 90$, hence $\triangle AMO \sim \triangle CNO$. Clearly the ratio of the areas is the ratio of the perpendiculars, and since the triangles are similar it is the ratio of $AO$ and $CO$ as well

Unless I royally messed something up I think this should be it.

@AshishAhuja How can I prove that the altitudes would be in opposite direction?

I mean what if they don't make vertical opposite angles.

I wanted to ask this only :-D

@RajdeepSindhu FYI Date sheet has been revised -_-

Now there is no preparatory leave for Sanskrit.

@Wolgwang Are you talking about $\angle MAO = \angle NCO$?

The altitudes do not make any "vertical opposite angles" with each other, at least according to the definition I found for it online.

The altitudes will be parallel to each other hence $\angle MAO = \angle NCO$, though you don't need this for the proof since two angles would satisfy the similarity condition.

I misread something :-/

@AshishAhuja I was talking about $\angle AOM = \angle CON$

@AshishAhuja This answers my query.

Thanks :-)

@Wolgwang They should be right? if they are in the same direction then they both will be perpendicular to DB(second diagram) and so they make a straight line. Thus, now we no longer similarity as base is common AO is actually AM and NC is OC lol. :P

so yeah maybe a more rigorous way to prove is to take both cases i.e. when they are opposite and when they are in same direction but no one will care in 10th grade lol

@Wolgwang Oh I see... I wonder if I missed a joke there lol

@HrishabhNayal Infact if they are on the same side then diagonals are perpendicular to each other and so the quadrilateral is a rhombus actually(right?).