No trigonometry allowed.
Let $\Delta ABC$ be inscribed inside a circle.Let $P$ be a point on the circle.Let $PD$ and $PE$ be perpendiculars on on $BC$ and $AC$ respectively.Let $DE$ when extended meet $AB$ at $F$. Then it is required to prove that $PF$ is perpendicular to $AB$.
For convenience of description,I will assume that $P$ is opposite to angle $A$.
I have tried the following things:
Drawing auxiliary lines: Unfortunately, apart from joining $P$ and $C$, $P$ and $A$, $F$ and $P$, and $P$ and $B$, I cannot see any obvious constructions. Moreover,the lines are unnecessarily cluttering up the diagrams.
Using Ptolemy's theorem or it's converse: Note that if $PF$ is indeed perpendicular to $AB$, this would make $BFPD$ a cyclic quadrilateral. Again, it would also make $AFPE$ a cyclic quadrilateral. Unfortunately, we do not have enough information regarding sides and diagonals.
Angle-chasing: Admittedly,this should have been my initial approach, but I was too busy with ratios. Let
$\angle BAC=x,\angle ACB=y$
Now we backtrack.If PF is indeed perpendicular to AB,we have that $\angle DPF=180-x-y=\angle ABC$.
Therefore,our task reduces to proving that $\angle DPF=ABC$
After that,I tried some other constructions(such as extending $FP$) but they turned out to be fruitless attempts.
The facts that I haven't used until now:
- The fact that $\Delta ABC$ is inscribed inside a circle.
The only way that I can see a use for this information is by constructing $PB,PA$ and saying that
$\angle BPA=\angle PAC$
But that does not help me get any further.
Some hints as to how I can approach such a problem will be appreciated.