$20-80-80$ triangle, rhombus with orthocenter, circumcenter Let $ABC$ triangle such that $\angle A=20^{\circ}$ and $\angle B=\angle C=80^{\circ}$.Let $D,E$ be point on lines $AC,AB$ respectively such that $BD,CE$ are angels bisector of triangle $ABC$.Let $H,O$ be orthocenter,circumecenter respectively of triangle $ABC$.Show that quadrilateral $DOEH$ is a rhombus. I have an ugly solution by trig bash that is not worth showing. However, I've noticed that we can place $A$ at the center and $B,C$ as vertices of a regular $18$-gon. This also gives nice collinearities for $D,E$. Or similarly, we can place $A,B,C$ as the vertices of a nonagon. Having seen solutions to variants of Langley’s Adventitious Angles proved concisely in this manner, I'm sure that we can do so for this question too.

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Let $\Gamma$ be circumcircle of $\triangle OHC$ and $S$ its circumcenter. Then $$\angle HSC = 2 \angle HOC = \angle BOC = 2\angle BAC = 40^\circ,$$ from which we get $$\angle SCH = \frac{180^\circ - \angle HSC}2=70^\circ = \angle ACH.$$ Thus $S$ lies on the line $AC$.

Let $OB$ intersect $\Gamma$ again at $X$. Since $\angle XOH = \angle HOC$, we have $XH=HC$, and obviously $HC=HB$, so $XH=HB$.

Furthermore, $$\angle HBX = \angle CBA - \angle CBH - \angle OBA = 80^\circ - 10^\circ - 10^\circ = 60^\circ.$$

Thus $\triangle XBH$ is equilateral. In particular $XB=BH$. Since $SX=SH$, we deduce that $$\angle HBS = \frac 12 \angle HBX = 30^\circ.$$

Thus $\angle CBS = 40^\circ$ and so $BS$ is the bisector of $\angle CBA$. Thus $S=D$.

We deduce that $OD=HD$. By symmetry, $OE=OD=HD=HE$. Therefore $DOEH$ is a rhombus.