geometry
For reference: In a triangle ABC; calculate the measure of the angle formed by BC and the straight line that passes through the orthocenter and the circumcenter($\measuredangle OEC=?$)
My progress…
I made the drawing and marked the found angles
H is orthocenter
C is circumcenter
Solutionby MathLover:
Let $P$ and $Q$ be, respectively, the orthocenter and circumcenter of $\triangle ABC$. Let $R$ be the intersection of the perpendicular bisector of $\overline{AB}$ with the altitude from $B$; and let $S$ be the intersection of the perpendicular bisector of ${AC}$ with the altitude from $C$.
Introduce $B^\prime = \overleftrightarrow{AB}\cap\overleftrightarrow{QS}$ and $C^\prime = \overleftrightarrow{AC}\cap\overleftrightarrow{QR}$. Then $\triangle ABC^\prime$ and $\triangle AB^\prime C$ are equilateral.
Observe that $R$ is orthocenter, circumcenter, and incenter for $\triangle ABC^\prime$; likewise, $S$ for $\triangle AB^\prime C$. Therefore, $R$ and $S$ lie on the bisector of $\angle A$. A little angle-chasing shows that $\triangle PRS$ and $\triangle QRS$ are equiangular, hence equilateral, so that $\square PSQR$ is a rhombus, and the result follows.
Hint:
$AH=2R\cos A$, $AI=4R\sin\frac{B}{2}\sin\frac{C}{2}$, $OA=R$, $\angle OAH=2\angle OAI=B-C$
$\triangle HOI=\triangle AHO-\triangle AHI-\triangle AIO$
Final answer $$2R^2\sin\dfrac{B-C}{2}\sin\dfrac{C-A}{2}\sin\dfrac{A-B}{2}$$
Best Answer
We have $ \small \angle ABC = 120^\circ$ in $ \small \triangle ABC$.
As $H$ is the orthocenter, $ \small \angle AHC = 180^\circ - \angle ABC = 60^\circ$. If you are not familiar with this, see the end of the answer where I show it.
$ \therefore \small AOCH$ is a cyclic quadrilateral ($ \small \angle AHC + \angle AOC = 180^\circ$).
So $ \small \angle CHO = \angle CAO = 30^\circ$
$ \small \angle CEO = \angle CHO + \angle HCE = 60^\circ$
If you want to show $ \small \angle AHC = 60^\circ$, note that as $ \small H$ is the orthocenter, $ \small AH$ will be perpendicular to $ \small BC$ so $ \small \angle BAH = 30^\circ$.
Say $ \small \angle ACB = x$. Then,
$ \small \angle GHC = 90^\circ - \angle GCH = 90^\circ - (30^\circ + x) = 60^\circ - x$
$ \small \angle GHA = 90^\circ - \angle GAH = 90^\circ - (60^\circ - x + 30^\circ) = x$
$ \small \angle AHC = \angle GHA + \angle GHC = 60^\circ$