Let $O'$ be the symmetric of $O$ wrt $AC$. It is well known that $AHO'O$ is a parallelogram, so $O'$ is on $HE$. By Sine Law, $BAC$ and $DAC$ have the same radius. As $\angle AO'C=\angle AOC=2\angle ABC=2\angle ADC$, we have that $O'$ is the circumcenter of $ADC$. Note that $OE=O'D$, so $EDOO'$ is an issoscelles trapezoid, so $BH=OO'=DE$.

Let $G,F$ be the feet of the perpendicular from $A$ and $B$ respectively. Let $O'$ be the reflection of $O$ with respect to $AC$. $\angle ADC=\angle B$ and $\angle AHC=\angle 180-\angle B \implies H,D,A,C$ cyclic. Now $O'A=O'D$ and $\angle AO'C=2\angle ADC$ which means $O'$ is the circumcircle of $\triangle ADC$. It is well known that $BH=OO'$ which implies $BHOO'$ is $\parallel gm \implies O'$ lies on $HE$. $BO=OE=O'H=O'D$ so $OD$ and $O'E$ have the same perpendicular bisector and this $\implies$ that $DE=O'O=BH$. Edit : very similar to #2 solution .

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Beautiful! Edit: Sniped by above Edit2: Also similar to #2 $\angle ADC=\angle ABC=180^{\circ}-\angle AHC$ $\implies$ $AHCD$ is cyclic, Let $F \in EH$ such that $BHFO$ is parallelogram, then $F$ is center of $\odot (AHCD)$ $\implies$ $OE=OC=CF=FD$ and hence, $OFED$ is an isosceles trapezium $\implies$ $BHED$ is also an isosceles trapezium $\implies $ $BH=DE$

GGPiku wrote: Let $O'$ be the symmetric of $O$ wrt $AC$. It is well known that $AHO'O$ is a parallelogram. I think you meant $BHO'O$.

Let $D' , O'$ the reflection of $D , O$ over $AC$ we have $\angle CD'A= \angle CDA=B \implies D' $lies on $(ABC)$ It's well-known fact $BH=2OM=OO'=2Rcos(b)$ thus $OO'HB$ is a parellelogram now we have $DO'=D'O=R=OE \implies OO'ED$ is a trapezium with equal diameter thus it's cyclic then $DE=OO'=BH$

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posting for fun: Let $O'$, $B'$ be the reflections of $O, B$ over $AC$, respectively. Therefore, we have $AB'DCH$ cyclic, since $$\measuredangle AHC =\measuredangle CBA=\measuredangle AB'C=\measuredangle ADC.$$ Since $\triangle AB'C$ is the reflection of $\triangle ABC$, then $O'$ is circumcentre of $(AB'DCH)$. Since $BH=2OM=OO'$, where $M$ is the midpoint of $AC$ and since $BH\perp AC\perp OO'$, we have that $OO'HB$ being parallelogram. We now want to show that $OEO'D$ is isosceles trapezoid. This is obvious, since $O'D=OE$ and $EO'\parallel OD$. Note that $OE$ cannot be parallel to $DO'$, since otherwise $OD=O'E<O'A=OX<OD$, where $X=OD\cap (ABC)$, easy to claim those inequalities, since $E$ must lie inside $(AB'C)$ and $D$ must lie outside $ABC$ (why?). We are done. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.882133056444922, xmax = 8.044799341843126, ymin = -5.217556006619065, ymax = 6.828027451329666; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((2.1709648279838567,1.533596002306361)--(-0.47,-1.88)--(4.55,-1.36)--cycle, linewidth(1) + zzttqq); /* draw figures */ draw((2.1709648279838567,1.533596002306361)--(-0.47,-1.88), linewidth(1) + zzttqq); draw((-0.47,-1.88)--(4.55,-1.36), linewidth(1) + zzttqq); draw((4.55,-1.36)--(2.1709648279838567,1.533596002306361), linewidth(1) + zzttqq); draw(circle((1.9807090002063497,-1.0476138096843763), 2.588212003202319), linewidth(1)); draw(circle((4.740255827777506,1.2212098119907375), 2.5882120032023193), linewidth(1)); draw((4.55,-1.36)--(7.286011611016771,0.754338453237164), linewidth(1)); draw((7.286011611016771,0.754338453237164)--(2.2895468275711575,0.38882362167511464), linewidth(1)); draw((7.286011611016771,0.754338453237164)--(3.7155014301756046,0.8731507772553995), linewidth(1)); draw((2.1709648279838567,1.533596002306361)--(6.030757470116876,3.464744274538606), linewidth(1)); draw((6.030757470116876,3.464744274538606)--(4.55,-1.36), linewidth(1)); draw((1.9807090002063497,-1.0476138096843763)--(4.740255827777507,1.2212098119907373), linewidth(1)); draw((-0.47,-1.88)--(3.7155014301756046,0.8731507772553995), linewidth(1)); draw((2.2895468275711575,0.38882362167511464)--(4.740255827777507,1.2212098119907373), linewidth(1)); draw((-0.47,-1.88)--(6.030757470116876,3.464744274538606), linewidth(1)); draw((-0.47,-1.88)--(7.286011611016771,0.754338453237164), linewidth(1)); /* dots and labels */ dot((2.1709648279838567,1.533596002306361),dotstyle); label("$A$", (2.236518003215782,1.6962788822583572), NE * labelscalefactor); dot((-0.47,-1.88),dotstyle); label("$B$", (-0.40360987797846487,-1.719386564036694), NE * labelscalefactor); dot((4.55,-1.36),dotstyle); label("$C$", (4.612633096290605,-1.1913609877978455), NE * labelscalefactor); dot((1.9807090002063497,-1.0476138096843763),linewidth(4pt) + dotstyle); label("$O$", (2.0385084121262134,-0.9108474004209572), NE * labelscalefactor); dot((2.2895468275711575,0.38882362167511464),linewidth(4pt) + dotstyle); label("$H$", (2.3520235980180306,0.524722134978412), NE * labelscalefactor); dot((6.030757470116876,3.464744274538606),dotstyle); label("$B'$", (6.097705029462368,3.626872395381647), NE * labelscalefactor); dot((7.286011611016771,0.754338453237164),linewidth(4pt) + dotstyle); label("$D$", (7.351765773029635,0.8877397186426204), NE * labelscalefactor); dot((3.7155014301756046,0.8731507772553995),linewidth(4pt) + dotstyle); label("$E$", (3.7875931334174022,1.0032453134448684), NE * labelscalefactor); dot((4.740255827777507,1.2212098119907373),linewidth(4pt) + dotstyle); label("$O'$", (4.810642687380173,1.3497620978516127), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]