Helpful lemma

[Phlembac Adib Hasan]

Here I'm giving a lemma that I use in many geometric problems.Many of you may know it.Actually it is for them who don't know it.
Acute angled $ \triangle ABC $ has orthocenter $H$ and circumcenter $O$.$ D,E,F $ are the feet of perpendiculars from $A,B,C$, respectively.Prove that $DE\perp CO. $

[itsnafi]

What's The Source?

[FahimFerdous]

@Nafi: do some angle chasing and you'll get it.

[*Mahi*]

@Nafi:
Quite a direct consequence if you can prove $O$ and $H$ are isogonal conjugates.

[itsnafi]

I've got it.But I just wanna know the related link to get more.
But after all,thnx to fahim vaia. .And to Mahi as well.

[Nadim Ul Abrar]

$(i)$ $ \angle BCO=\angle ACH$ . $(ii)$ $ ABDE$ is cyclic .

Using $(i)$ $(ii)$ we can prove that $\triangle DCI$ ~ $\triangle CEH$ ( $ I$ is the intersecting point of $OC$ and $DE$)

[nafistiham]

Proof:

\[\angle EDA=\angle EBA=\frac{\pi}{2}-\angle A=\frac{\pi-2\angle A}{2}=\angle OCD\]
\[\angle OCD+\angle EDC=\angle EDA+\angle EDC=\angle ADC=\frac{\pi}{2}\]

কিন্তু আদিব $H$ এর দরকার কি ? বুঝলাম না

[Nadim Ul Abrar]

Proof:

\[\angle EDA=\angle EBA=\frac{\pi}{2}-\angle A=\frac{\pi-2\angle A}{2}=\angle OCD\]
\[\angle OCD+\angle EDC=\angle EDA+\angle EDC=\angle ADC=\frac{\pi}{2}\]

কিন্তু আদিব $H$ এর দরকার কি ? বুঝলাম না

ওই মিয়া পোলার মনে এম্নেও দুঃখ , আর উনি আরও দুঃখ দেয় ...

[Tahmid Hasan]

can be solved using complex too.
taking co-ordinates $A=a,B=c,C=c,O=0$ and inscribing $\triangle ABC$ in an unit circle,
we get $2D=a+b+c- \frac {bc}{a},2E=a+b+c- \frac {ac}{b}$
which gives straightforward result

[Niloy Da Fermat]

I've got it.But I just wanna know the related link to get more.
But after all,thnx to fahim vaia. .And to Mahi as well.

আমি plane euclidean geometry বই থেকে এইসব পরসি। চমতকার আছে ঐখানে, দেখতে পারো। আর মাহি, isogonal conjugate জিনিসটা কি?