Since $E$ belongs to $K$ and to the circumcircle of $\triangle BCD$, then $$ \measuredangle BDE=\measuredangle DEB=\measuredangle DCB $$Also, since $H$ is the orthocenter of $\triangle ABC$, then $AH\perp BC$, and so $$ \measuredangle HAD+\measuredangle DCB=90^o $$Now, since $\triangle BAD$ is isosceles in $B$ and $BH\perp AD$, then $BH$ is the perpendicular bisector line of segment $AD$, and so $$ \measuredangle ADH=\measuredangle HAD $$And, since $HF\perp AD$, then $$ \measuredangle ADH+\measuredangle DHF=90^o $$Combining these equalties, $$ \measuredangle BDE=\measuredangle DEB=\measuredangle DCB=90^o-\measuredangle HAD=90^o-\measuredangle ADH=\measuredangle DHF $$Therefore, $$ \measuredangle BDE=\measuredangle DHF \, , $$and the heading is proved.

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DylanN wrote: Let $ABC$ be a triangle with $\angle ABC \neq 90^\circ$ and $AB$ its shortest side. Denote by $H$ the intersection of the altitudes of triangle $ABC$. Let $K$ be the circle through $A$ with centre $B$. Let $D$ be the other intersection of $K$ and $AC$. Let $K$ intersect the circumcircle of $BCD$ again at $E$. If $F$ is the intersection of $DE$ and $BH$, show that $BD$ is tangent to the circle through $D$, $F$, and $H$. $AB=BD=BE$ and $BDCE$ is a cyclic quadrilateral Let $A'$ be the foot of the height from $A$ to $BC$ and $B'$ be the foot of the height from $B$ to $AC$: Let $\angle BAA'=\beta, \angle A'AC=\alpha$ and $\angle BCA=\theta$ It is easy to see that $\angle ABC=90^{\circ}-\beta$ Since $BA=BD \Rightarrow \angle BDH=\beta$ and $\angle HDA=\alpha$ Since $BD=BE \Rightarrow \angle BED=\theta$ Since $BDCE$ is a cyclic quadrilateral $\Rightarrow \angle BCD=\theta$ In $\triangle ABC: \alpha+\beta+90^{\circ}-\beta+\theta=180^{\circ}$ $\Rightarrow \alpha+\theta=90^{\circ}$ In $\triangle B'FD:$ $90^{\circ}+\angle B'FD=\alpha+\beta+\theta$ $\Rightarrow \angle B'FD=\alpha + \beta + \theta - 90^{\circ}$ $\Rightarrow \angle B'FD=\beta=\angle BDH$ $\Rightarrow BD$ is tangent to $\odot(DFH) _\blacksquare$