Connect $OE$ and $OF$. So $\angle{OEA}=\angle{OAE}$, $\angle{OFB}=\angle{OBF}$. And $\angle{OEP}=\angle{OFP}=90^{o}$. Since the sum of the angles of a pentagon is $540^{o}$, so: \[\angle{A}+\angle{B}+\angle{BFO}+\angle{OFP}+\angle{FPE}+\angle{PEO}+\angle{OEA}=540^o\] Therefore: $2\angle{A}+2\angle{B}+\angle{FPE}=360^{o}$ Since $2\angle{A}+2\angle{B}+2\angle{C}=360^o$, so $\angle{FPE}=2\angle{C}$. Now since $PE$ and $PF$ are tangents, so $PE=PF$. Therefore $P$ is the orthocentre of $\triangle{CEF}$. Hence: $\angle{A}+\angle{ACP}=\angle{OEA}+\angle{CEP}=90^{o}$, which means $CP \perp AB$.

Attachments:

P is the circumcenter not the orthocenter

marko avila wrote: P is the circumcenter not the orthocenter It was a typo... Thanks for pointing it out!

After two years... Let $H$ be the foot of altitude from $C$. Since $PE = PF$, $\measuredangle PEF = \measuredangle PFE$. $\measuredangle CEP = 90^{\circ}-\measuredangle CAH = \measuredangle HCE$. Also $\measuredangle CFP = \measuredangle HCF$. Consider the cevians $CH, FP, EP$ of $\Delta CEF$. By the trigonometric Ceva they concur. Hence $P$ lies on $CH$.

Or very short with projective geometry: Call $EF \cap AB = {G}$. Now, the polar of $G$ goes through $C$ and is perpendicular to $AB$, so it is the altitude through $C$. Also, we know that $EF$ is the polar of $P$, and $G$ lies on it. So $P$ lies on the polar of $G$, $\mathbb{QED}$

Denote $H=AF \cap BE$, of course, $H$ is the ortochenter of $\triangle{ABC}$. Using Pascal's Theorem for the cyclic hexagon $AEEBFF$ we get $C-P-H$ collinear, qed.

Quote: PP. Let $\triangle{ABC}$ be an acute triangle with the orthocenter $H$ . The circle $w$ with the diameter $[BC]$ intersects $AC$ and $AB$ at points $E$ and $F$ respectively, i.e. $H\in BE\cap CF$ . The tangents drawn to the circle $w$ through $E$ and $F$ intersect at $P$ . Show that $P\in AH$ . Proof. Denote the midpoints $M$ , $N$ of the segments $[BC]$ , $[AH]$ respectively. $\{\begin{array}{ccc} NF=NH & \implies & m(\widehat{NFC})=B\\\ NE=NH & \implies & m(\widehat{NEB})=C\end{array}$ . Since $\{\begin{array}{c} m(\widehat{PFC})=B\\\ m(\widehat{PEB})=C\end{array}$ obtain that $P\in FN\cap EN\implies$ $P\equiv N\in AH$ .

$ABFE$ is cyclic, $\widehat{FEC}=\widehat{ABC}$. Easy angle chasing gives $\widehat{EOF}=180^\circ- 2\cdot m(\angle BCA)$, i.e. $EO, FO$ are tangent to circle $\odot (CEF)$. $PEOF$ cyclic gives $\widehat{PEF}=\widehat{PFE}=\frac{\widehat{FEO}}{2}=90^\circ-m(\angle ACB)$, hence $P$ is the circumcenter of $\odot (CEF)$, or $\widehat{FCP}=90^\circ-\widehat{CEF}=90^\circ-\widehat{ABC}$, $\equiv CP\bot AB$. Best regards, sunken rock

Clearly $BE$ and $CF$ are the altitudes of triangle $ABC$.suppose the common point of altitudes is $H$.Again $CEHF$ is cyclic with diameter $CH$,this is enough to prove that $P$ is the circumcenter of $CEHF$,now supposing $X$ is the midpoint of $CH$, and extended $XE$ to $G$.now $XE=XC$.......(1) ${\angle}ABE={\angle}AFE$,${\angle}HFE={\angle}HCE$,now using (1) we get ${\angle}AEG={\angle}ABE$,so $XG$ is tangent to circumcircle of $EABF$,hence $X$ and $P$ are not distinct,hence we are done.