Call F the symmetric of A w.r.t. D, from Balkan $ E \in DF$, so $ \angle DHE = \angle AHD = \angle ABC$

Litlle 1000t wrote: Let $ ABC$ be an acute triangle with circumcircle $ w = C(O)$ and orthocenter $ H$ . Denote $ \{\begin{array}{c} D\in AB\ ,\ CD\perp AB \\ \ E\in BC\ ,\ DE\perp DO\end{array}$ . Prove that $ \widehat{DHE}\equiv\widehat{ABC}$ . Proof. The circle $ w$ cut again the line $ CD$ in the point $ F$ . Denote $ K\in AF\cap DE$ . Apply the "property of butterfly" (see here ) to the interior point $ D\in CF\cap AB\cap EK$ of the circle $ w$ : $ \boxed {\ OD\perp \overline {EDK}\ \Longleftrightarrow\ DE = DK\ }$ . Thus, $ \{\begin{array}{c} DE = DK \\ \ DH = DF\end{array}\|$ $ \implies$ $ HE\parallel FA$ $ \implies$ $ \widehat {DHE}\equiv\widehat {DFA}\equiv\widehat {CBA}$ , i.e. $ \widehat {DHE} = \widehat {ABC}$ .

Virgil Nicula wrote: Litlle 1000t wrote: Let $ ABC$ be an acute triangle with circumcircle $ w = C(O)$ and orthocenter $ H$ . Denote $ \{\begin{array}{c} D\in AB\ ,\ CD\perp AB \\ \ E\in BC\ ,\ DE\perp DO\end{array}$ . Prove that $ \widehat{DHE}\equiv\widehat{ABC}$ . Proof. The circle $ w$ cut again the line $ CD$ in the point $ F$ . Denote $ K\in AF\cap DE$ . Apply the "property of butterfly" (see here ) to the interior point $ D\in CF\cap AB\cap EK$ of the circle $ w$ : $ \boxed {\ OD\perp \overline {EDK}\ \Longleftrightarrow\ DE = DK\ }$ . Thus, $ \{\begin{array}{c} DE = DK \\ \ DH = DF\end{array}\|$ $ \implies$ $ HE\parallel FA$ $ \implies$ $ \widehat {DHE}\equiv\widehat {DFA}\equiv\widehat {CBA}$ , i.e. $ \widehat {DHE} = \widehat {ABC}$ . Dear Virgil , to which form of the buterfly theorem are you reffering to ? The form you give at the link , is the classic form . Do you use any generalisation of the theorem ? Babis

It is shortlisted problem. (1996, G3)

stergiu wrote: Virgil Nicula wrote: Litlle 1000t wrote: Let $ ABC$ be an acute triangle with circumcircle $ w = C(O)$ and orthocenter $ H$ . Denote $ \{\begin{array}{c} D\in AB\ ,\ CD\perp AB \\ \ E\in BC\ ,\ DE\perp DO\end{array}$ . Prove that $ \widehat{DHE}\equiv\widehat{ABC}$ . Proof. The circle $ w$ cut again the line $ CD$ in the point $ F$ . Denote $ K\in AF\cap DE$ . Apply the "property of butterfly" (see here ) to the interior point $ D\in CF\cap AB\cap EK$ of the circle $ w$ : $ \boxed {\ OD\perp \overline {EDK}\ \Longleftrightarrow\ DE = DK\ }$ . Thus, $ \{\begin{array}{c} DE = DK \\ \ DH = DF\end{array}\|$ $ \implies$ $ HE\parallel FA$ $ \implies$ $ \widehat {DHE}\equiv\widehat {DFA}\equiv\widehat {CBA}$ , i.e. $ \widehat {DHE} = \widehat {ABC}$ . Dear Virgil , to which form of the buterfly theorem are you reffering to ? The form you give at the link , is the classic form . Do you use any generalisation of the theorem ? Babis Well , I had the extension of side BC and not side BC. Everything is OK now. Virgil , sorry ! Babis

Sorry,Why $ \boxed {\ OD\perp \overline {EDK}\ \Longleftrightarrow\ DE = DK\ }$ ? Thank you

Can anyone,please,explain to me why $ OD\perp \overline {EDK}\ \Longleftrightarrow\ DE = DK$ Thanks

Let $F= (BHC) \cap \overline{AB}$ and let $G_1,G_2$ be the points of intersection $\overline{ED}$ with the circumcircle. Throw the configuration onto the complex plane and scale down the circumcircle to the unit circle and let $b=\frac{1}{a}$. Using the fact that $CA=CF$ we have that $a+f=2d$, where $d=\frac{1}{2}\left(a+\frac{1}{a}+c-\frac{1}{c}\right)$, this implies that $f=\frac{1}{a}+c-\frac{1}{c}$. By definition we have that: $$\frac{g_1-g_2}{\overline{g_1-g_2}}=-\frac{d}{\overline{d}}$$$$g_1=\frac{d}{g_2.\overline{d}}$$ Plugging in what we have we get that $HFG_1G_2$ is cylic. Now by the theory of the radical center on the circles $(ABC),(BCHF)$ and $(HFG_2G_1)$ we have that the radical center is $E$, this implies that $\angle EHD = \angle FHD = \angle ABC$

Here is a solution by mueller.25 Switch to $A$ labelling. Observe that $O,H$ are isogonal in angles $A,C,D$ and so $H$ has an isogonal conjugate in $ACDE$. So, $\angle EHD + \angle AHC = 180 \implies \angle DHE = \angle ABC$. $\blacksquare$

Sorry I am stupid but can someone post the problem wording in $A$ labelling? I do not know which side the line through point $ D$ and perpendicular with $OD$ meets?

Oops, should probably have been specific. Rephrased Problem wrote: Let $ABC$ be a triangle with $H$ orthocenter and $O$ circumcenter. Let $AH$ meet $BC$ at $D$ and let $E$ be a point on $AB$ such that $\angle EDO = 90^\circ$. Show that $\angle DHE = \angle ABC$

Shishkin wrote: It is shortlisted problem. (1996, G3) Indeed, see ISL 1996 G3 .