Let $C_1\equiv CH\cap AB$ $\Longrightarrow$ $C_1\in \odot (BDC)$ and $C_1\in \odot (ACA_1)$ $\Longrightarrow$ $\measuredangle C_1DB=\measuredangle C_1CB=\measuredangle C_1AA_1$ $\Longrightarrow$ $BD$ is tangent to $\odot (ADC_1)$ $\Longrightarrow$ $BD^2=BC_1.BA...(1)$. Let $X\in \odot (AHC_1)$ such that $BX$ is tangent to $\odot (AHC_1)$ $\Longrightarrow$ $BX^2$ $=$ $BC_1.BA$ $...(2)$ $\Longrightarrow$ by $(1)$ and $(2)$ we get $BX=BD$.

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Problem: Let $\Delta DEF$ be the orthic triangle WRT $\Delta ABC$ with orthocenter $H$. Let $M$ be midpoint of $\overline{BC}$. Let $N$ $\equiv$ $\odot (BFEC)$ $\cap$ $\overline{AD}$. Let $BL,BK$ be tangents from $B$ to $\odot (AEF)$. Show, $BK=BN=BL$ Proof: Let $N' \in \overline{AD}$, such $BK=BN'=BL$. Perform inversion $\Psi$ around $\odot (KLN')$. Since, $\Psi (F)$ $\equiv$ $A$ and $\Psi ( E) $ $\equiv$ $H$ $\implies$ $\Psi ( \odot (BFEC)) $ $\equiv$ $AH$ $\implies$ $N' \in \odot (BFEC)$ $\implies$ $N \equiv N'$

Hehe,this was so simple and direct,but still I will write the solution. From similarity ,we have $BD^2=BA_{1}.BC$. From power of $B$ w.r.t $\odot(AC_{1}A_{1}C)$ $BA_{1}.BC=BC_{1}.BA=BD^2$. Hence we are done.