If a line is perpendicular to two distinct lines in a plane, then it is perpendicular to any line in this plane.
Is there any way to prove this result from Stewart Theorem? I started out of contradiction, but then I didn't understand where to start applying Stewart's theorem. By the way, I don't see a relationship with Stewart's theorem either.
Could anyone give a hint to this problem?
Choose points $B$ and $C$ on lines $s$ and $s'$ such that segment $BC$ intersects line $t$ at $D$. In addition, set $PB=b$, $PC=c$, $CD=n$, $BD=m$, $PD=d$, $AD=d'$, $AP=h$.
From Stewart's theorem applied to triangle $PBC$ one gets: $$ b^2m+c^2n=(m+n)(d^2+mn). $$ From Stewart's theorem applied to triangle $ABC$, taking into account that $AB=\sqrt{b^2+h^2}$, $AC=\sqrt{c^2+h^2}$, one gets: $$ (b^2+h^2)m+(c^2+h^2)n=(m+n)(d'^2+mn). $$ Subtracting the previous equality we thus get: $$ h^2=d'^2-d^2 $$ which entails $\angle APD=90°$.