Find congruent triangles We have $AH = a \cot A = a$. Thus $AX = \frac{a}{\sqrt{2}}$. Also, $AO = \frac{a}{2\sin A} = \frac{a}{\sqrt{2}}$ giving $AX=AO$. We also have $\angle HAX = 45^{\circ} = \angle BAC$ giving $\angle OAD = \angle XAD$. Thus we have $\triangle ADO \cong \triangle ADX$ and we are done.

Let $X'$ be the Midpoint of arc $AH$ of the circumcircle of triangle $ADH$ that doesn't contain $D$,and let $F$ be the foot of altitude from $C$. It is easy to see that $DO$ is the perpendicular bisector of $AB$ $\Longrightarrow \angle{ADO} =45^{\circ} $ and we have $\angle{ADX'} =45 ^{\circ}$ $\Longrightarrow $ $D,O$ and $X'$ are collinear $\Longrightarrow \angle{XDO}=90 ^{\circ}$. In the same way we can prove that $F,O$ and $X$ are collinear $\Longrightarrow \angle{DXO} =\angle{DXF} =\angle{DAF} =45 ^{\circ} $. $\Longrightarrow \angle{DXO}=\angle{DOX}=45^{\circ} \Longrightarrow DX=DO$. So we are done. $\blacksquare$

Notice that $DXAH\stackrel{+}{\sim}DOBC$ hence $\triangle DXO\sim\triangle DAB$. But clearly $\triangle DAB$ is isosceles hence we are done.

Let $O'$ be the reflection of $O$ over $AC$. We see that $O'H=OA=O'A$. Hence, $O'HA$ is isosceles. Now, since $O,H$ are isogonal conjugates, we get $\angle O'AH = 45^\circ$. So, $O' \in \odot (AH)$. Hence, $X\equiv O'$. So, $DX=DO$. $\blacksquare$.

See that $BCDO\cong AHDX$! Best regards, sunken rock

Let $O'$ be the reflection of $O$ in $AC$. Note that $O$ and $H$ are isogonal conjugates. \[\angle O'AH = \angle O'AD + \angle DAH = \angle DAO + \angle OAB = \angle CAB = 45^{\circ}.\]It's well-known that the reflection of $H$ in $\overline{AC}$ lies on $\odot(ABC)$, so, $O'$ is the centre of the $\odot(AHC)$, so, $\angle O'HA = \angle OHA' = 45^{\circ}$. Therefore, $\angle AO'H = 90^{\circ}$ which implies $O' \in \odot(ADH)$. However, since $O'A = O'H$ must be the midpoint of arc $\widehat{ADH}$, thus, $X \equiv O'$.

a solution by Stelios Karpathios from here

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It also can be solved by using inversion with radius bc/2 through A

Note that in any triangle $ABC$ with orthocentre $H$ and circumradius $R$ the relation $AH=2R.cos\angle{A}$ satisfies. since $\angle{HAX}=\angle{BAC}=45^{\circ}$ hence $\angle{DAX}=\angle{BAH}=\angle{DAO}$. on other hand: $AX=\frac{AH}{\sqrt{2}}=\frac{2R.cos\angle{45^{\circ}}}{\sqrt{2}}=R=AO$. Hence $AD$ is the perpendicular bisector of $OX$ and we are done.

The condition implies $\triangle DOB \cong \triangle DOA \cong \triangle DXA$ so $DX=DO$

Let $AB\cap XH=T,EX\cap AC=K$ and $E$ be the altitude from $C$ to $AB$. $\angle BAC=45=\angle HAX \implies 90-\angle B=\angle BAH=\angle DAX=\angle DHX=\angle BHT$ which means $TH$ is tangent to $(ABH)$. \[TH^2=TB.TA\]\[TH.TX=TE.TA\]By dividing these two, we get \[\frac{TH}{TX}=\frac{TB}{TE}\]$\implies BH\parallel EX$ $\implies EX\perp AC$ $EO \perp AC$ and $EX\perp AC$ gives $E,O,X$ are collinear. $\angle OAK=90-B=\angle KAX$ and $AK\perp OX$ $\implies OK=KX$ $\implies DO=DX$

Let $E$ be the foot of altitude from $C$ to $AB$ $Claim$ $1:$ $E,O,X$ are collinear and $BH\parallel EX$ $Proof:$ $EX$ is a angle bisector of $\angle AEH$, so $EX\perp AC$($\triangle AEC$ is an isosceles right triangle) It means that $EX$ is the perpendicular bisector of $AC$, meaning that $E,O,X$ are collinear. Since, $EX\perp AC$ and $BH\perp AC$ , $\implies BH\parallel EX$ $Claim$ $2:$ $DO=OX$ $Proof:$ $\angle EXD= \angle EAD= \angle BAC=45$ and $\angle DOX= \angle DCE=45$(Since B, E, O, D, C are concyclic) $\implies DO=DX$ and we are done.

Take $M$ as mid point of $AH$.By simple angle chasing, prove that $/Delta AXM$ is isosceles right angled triangle. Now, take $O'$ as reflection of $O$ over $CA$. Using angle chasing once more, find that $/Delta AO'M$ is also isosceles right triangle. Hence $X \equiv O'$. $\implies DO=DX$.