Let $ABC$ be an obtuse triangle, with $AB$ being the largest side.
Draw the angle bisector of $\measuredangle BAC$. Then, draw the perpendiculars to this angle bisector from vertices $B$ and $C$, and call their feet $P$ and $Q$, respectively.
$D$ is the point in the line $BC$ such that $AD \perp AP$.
Prove that the lines $AD$, $BQ$ and $PC$ are concurrent.
WLOG let $AC<AB$, so that $D$ lies closer to $C$ than to $B$. Let $X=CP\cap AD$ and $E=AQ\cap BC$. Then it suffices to prove that $X$, $D$, and $B$ are collinear, which is equivalent to $\triangle XCQ\sim\triangle APB$. Remark that \[\dfrac{XC}{XP}=\dfrac{DC}{DB}\qquad\text{and}\qquad\dfrac{QC}{BP}=\dfrac{CE}{EB}.\] Thus if we can prove $\frac{DC}{DB}=\frac{CE}{EB}$, then we are done, since we already know $\angle ACQ=\angle APB$ from $QC\parallel BP$. But this is trivial: remark that $\angle CAE=\angle BAE$ and $DA\perp AE$ are precisely two conditions that make $D$, $C$, $E$, and $B$ form a harmonic division in that order, from which the desired ratio equality is a natural consequence.