Problem
The altitudes $AD$, $BE$ and $CF$ of $\Delta ABC$ meet at the orthocenter $H$.
Since $\angle AEH=\angle AFH=90^\circ$, $AEHF$ is a cyclic quadrilateral.
Let $U$ and $V$ be points lying on the circle $AEHF$ such that $BU$ and $CV$ are the tangents to the circle $AEHF$ at $U$ and $V$ respectively.
Since $\angle BEC=\angle BFC=90^\circ$, $BCEF$ is a cyclic quadrilateral.
Let $W$ be the intersection point of the circle $BCEF$ and the line segment $AH$.
Prove that $BU=BW$ and $CV=CW$.
My Progress
Since $\angle CDH=\angle CEH=90^\circ$, $CDHE$ is a cyclic quadrilateral.
Considering the power of point with respect to circles $AEHF$ and $CDHE$, we have $BU^2=BE\cdot BH=BC\cdot BD$.
Since $\angle BDH=\angle BFH=90^\circ$, $BDHF$ is a cyclic quadrilateral.
Considering the power of point with respect to circles $AEHF$ and $BDHF$, we have $CV^2=CF\cdot CH=BC\cdot CD$.
Hence, we have $BU^2+CV^2=BC\cdot BD+BC\cdot CD=BC\cdot (BD+CD)=BC^2$
Since $BC$ is a diameter of the circle $BCEF$, we have $BW^2+CW^2=BC^2$.
Hence, any one of the equalities $BU=BW$ and $CV=CW$ implies the other.
First, let's investigate the length of $WD$.
We know that:
$$AF.AB=AW.AM=AW.(AW+WM)=AW.(AW+2WD)\\=(AD-WD)(AD-WD+2DW)=AD^2-WD^2 \\ \implies WD^2=AD^2-AF.AB.$$ Now, note that:
$$BW^2=BD^2+WD^2=BD^2+AD^2-AF.AB=AB^2-AF.AB \\=BF.AB=BU^2.$$
We are done.