Angle chase to obtain $E, Z, H$ collinear Thus $\angle{ADH}=\angle{AZH}=\angle{BZE}=\angle{BDE}$.Since $BD=AD$,$\angle{DAH}=\angle{DBE} \therefore BDE$ is congruent to $ADH$ and the conclusion follows $\blacksquare$

Note that $\angle{DZE} = \angle{DBE} = \angle{DAH}$. So, $E,Z,H$ are collinear. Also, note that $\angle{ADH}=\angle{AZH}=\angle{BZE}=\angle{BDE}$. Since $AD = DB$, $\Delta EDB \cong \Delta ADH$. Thus, $AH = BE$. $\blacksquare$.

Attachments:

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(16cm); 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 = -15.778341463414641, xmax = 8.026536585365854, ymin = -7.04, ymax = 6.150243902439026; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.08,4.74)--(-8.1,-3.64)--(5.9,-3.7)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.08,4.74)--(-8.1,-3.64), linewidth(2) + rvwvcq); draw((-8.1,-3.64)--(5.9,-3.7), linewidth(2) + rvwvcq); draw((5.9,-3.7)--(-4.08,4.74), linewidth(2) + rvwvcq); draw(circle((-1.0921893617962388,-1.8475177524557416), 7.233422602660729), linewidth(2)); draw((xmin, -0.004285714285714289*xmin-3.6747142857142854)--(xmax, -0.004285714285714289*xmax-3.6747142857142854), linewidth(2)); /* line */ draw((xmin, 0.9787477138879433*xmin + 8.73329067266281)--(xmax, 0.9787477138879433*xmax + 8.73329067266281), linewidth(2)); /* line */ draw(circle((-10.349521278202536,-0.9333051174741805), 3.5194521119312636), linewidth(2)); draw(circle((-4.095083227675184,1.220580209122481), 3.519452111931268), linewidth(2)); draw((-7.614015116547947,1.2810907838332637)--(-8.1,-3.64), linewidth(2)); draw((-7.614015116547947,1.2810907838332637)--(-6.830589389329778,-0.993815692184963), linewidth(2)); draw((-6.830589389329778,-0.993815692184963)--(-0.6270262736970033,1.8198498747497704), linewidth(2) + dtsfsf); draw((-7.614015116547947,1.2810907838332637)--(-0.6270262736970033,1.8198498747497704), linewidth(2)); draw((-12.622159738178468,-3.620619315407807)--(-6.830589389329778,-0.993815692184963), linewidth(2) + dtsfsf); draw((-7.614015116547947,1.2810907838332637)--(5.9,-3.7), linewidth(2)); /* dots and labels */ dot((-4.08,4.74),dotstyle); label("$A$", (-4.461268292682931,5.194146341463417), NE * labelscalefactor); dot((-8.1,-3.64),dotstyle); label("$B$", (-8.42224390243903,-4.269268292682929), NE * labelscalefactor); dot((5.9,-3.7),dotstyle); label("$C$", (6.1338536585365855,-4.249756097560978), NE * labelscalefactor); dot((-7.614015116547947,1.2810907838332637),dotstyle); label("$D$", (-8.26614634146342,1.1941463414634148), NE * labelscalefactor); dot((-12.622159738178468,-3.620619315407807),dotstyle); label("$E$", (-12.910048780487811,-4.366829268292685), NE * labelscalefactor); dot((-6.830589389329778,-0.993815692184963),dotstyle); label("$Z$", (-6.353951219512199,-1.4204878048780496), NE * labelscalefactor); dot((-0.6270262736970033,1.8198498747497704),dotstyle); label("$H$", (-0.207609756097563,2.0136585365853663), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $$\angle DZH=180^{\circ}-\angle DAC=\angle DBC=180^{\circ}-\angle DBE=180^{\circ}-\angle DZE \implies E-Z-H$$Now, $\angle EDB=\angle EZB=\angle AZH=\angle ADH$ and since, $BD=AD$, hence, $\Delta BED \cong \Delta AHD$ $\implies$ $BE=AH$

$\measuredangle DZE = \measuredangle DBE \measuredangle DBC = \measuredangle DAC = \measuredangle DZH$ Hence, $E,Z,H$ are collinear. $\measuredangle EDB = \measuredangle EZB = \measuredangle HZA$ and $BD = AD$. So $\Delta DBE \cong \Delta DAH \Rightarrow BE = AH$

We have: $\widehat{AHZ}$ = $\widehat{EDZ}$ = $\widehat{ZBC}$ or $B$, $C$, $H$, $Z$ lie on a circle We also have: $\widehat{HZE}$ = $\widehat{HZD}$ + $\widehat{DZE}$ = $180^o$ $-$ $\widehat{DAH}$ + $\widehat{DBE}$ = $180^o$ or $H$, $Z$, $E$ are collinear But: $\widehat{DEZ}$ = $\widehat{DBZ}$ = $\widehat{DAZ}$ = $\widehat{DHZ}$ then $\triangle$ $ABD$ $\sim$ $\triangle$ $HED$ or $DH$ = $DE$ Combine with: $AD$ = $BD$ and $\widehat{BDE}$ = $\widehat{BZE}$ = $\widehat{AZH}$ = $\widehat{ADH}$, we have: $\triangle$ $DEB$ = $\triangle$ $DHA$ or $BE$ = $AH$

By Miquel's theorem,$C,B,Z,H$ are concyclic.$BD=AD,\angle EBD=\angle DAH,\angle EDB=\angle ACB=\angle AZH=\angle ADH \implies \bigtriangleup EDB \cong \bigtriangleup ADH \implies BE=AH$.

$DA=DB$ $\angle AHD=\angle AZD=\angle DEB$ $\angle DAH=\angle DAC=\angle EBD$ So $\triangle DAH \cong \triangle DBE$ and we are done

Killed this in 15 minutes just a few months ago, now I can kill this in 1 minute. $\measuredangle BED=\measuredangle BZD=\measuredangle AZD=\measuredangle AHD$ $\measuredangle EBD=\measuredangle CBD=\measuredangle CAD=\measuredangle HAD$, so $\triangle AHD\sim \triangle BED$. With $AD=BD$ we have $\triangle AHD\cong \triangle BED$ so $AH=BE$.