Dear Mathlinkers, 1. X the point of intersection of AB and PC 2. according to Appolonius, (C,D,X, P) = -1 3. the pincil (A; C, D, X, P) = -1 4. according to Pappus (PA // EF), B is the midpoint of EF. Sincerely Jean-Louis

I have another projective solution. Let the tangents to $ \Gamma $ at $C,D$ intersect at $Q$. Then the $Q$ lies on the line $AB$, $ACBD$ is harmonic. $CDFE$ is cyclic and $CD$, $EF$ are inversion parallel. Hence $AB$ is symmedian for $ACD$ $ \Rightarrow $ $ AB $ is median for $AEF$.

∠PAE = ∠AEB = ∠ABC so triangles ABC and ABE are similar. so we have: (AB)/(AE) = (BC)/(BE) so : BE = (BC)×(AE)/(AB) ∠180-DAP = ∠AFE = ∠ABD so triangles ABD and ABF are similar so we have: (BD)/(BF) = (AD)/(AB) so : BF = (BD)×(AB)/(AD) So we have to prove: (BD)×(AB)/(AD) = (BC)×(AE)/(AB) Or : (AB^2) = (AD)×(BC)×(AE)/(BD) But we have : ∠ABC = ∠AEB so AB is tangent to the circumcircle of triangle BCE. So : (AB^2) = (AC)×(AE) So we have to prove that: (AB^2) = (AC)×(AE) =? (AD)×(BC)×(AE)/(BD) Or: (AC)/(AD) =? (BC)/(BD) But triangles ACP nd APD are similar so : (AC)/(AD) = (PA)/(PD) Triangles BCP and BDP are similar too so : (BC)/(BD) = (PB)/(PD) But we have PA = PD so : (BC)/(BD) = (PB)/(PD) = (PA)/(PD) = (AC)/(AD) And we are done.

Firstly $\angle{CDA}=\angle{EAP}=\angle{FEA}$ so $CDFE$ is cyclic. It is well known that $ABCD$ is harmonic, so $AB$ is the symmedian of $CAD$. However, since $CDFE$ is cyclic triangles $CAD$ and $FAE$ are isogonal to one another so the symmedian of $CAD$ is the median of $FAE$, hence $B$ is the midpoint of $EF$.

This is my first ever proof using Projective Geometry So, I maybe wrong, So I request to please check my solution... Let $DC\cap AB=Q$. Now as $CDAB$ is a harmonic quadrilateral. Hence, $(C,D;A,B)=-1$. Now, $-1=(C,D;A,B)\overset{A}{=} (C,D;Q,P)$. Now, $(C,D;Q,P)\overset{A}{=} (E,F;B,P_{\infty})=-1$. Now by Lemma 9.8 from EGMO we get $B$ is the midpoint of $EF$. @below Ya I thought that, Thanks.

@above It's fine. You may also $CDAB$ project on $EF$ without extra step: $-1=(C,D;A,B)\overset{A}{=} (E,F;A_{\infty},B)$.

We use Law of Sines, $$\frac{EB}{\sin{\angle{EAB}}}=\frac{AE}{\sin{\angle{ABE}}}\quad (1)$$and $$\frac{BF}{\sin{\angle{BAF}}}=\frac{AF}{\sin{\angle{ABF}}} \quad (2)$$Dividing $(1)$ by $(2)$ we obtain that $$\frac{EB}{BF}=\frac{AE\cdot \sin{\angle{EAB}}}{AF\cdot \sin{\angle{BAF}}} \quad (3)$$We obviously have that $\angle{BAF}=\angle{BAD}=\angle{BCD}$ and $\angle{EAB}=\angle{CAB}=\angle{CDB}$. By Law of Sines, $$\frac{BD}{\sin{\angle{BCD}}}=\frac{CB}{\sin{\angle{CDB}}}\Longleftrightarrow \frac{\sin{\angle{CDB}}}{\sin{\angle{BCD}}}=\frac{CB}{BD}\quad (4)$$By $(3)$ and $(4)$, we have that $$\frac{EB}{BF}=\frac{AE\cdot CB}{AF\cdot BD} \quad (5)$$We also obviously have $\triangle PBD\sim \triangle PCB$ and $\triangle PCA\sim \triangle PAD$: $$\frac{CB}{BD}=\frac{PC}{PB} \text{ and } \frac{PC}{PA}=\frac{AC}{AD},$$hence $(5)$ takes the following form: $$\frac{EB}{BF}=\frac{AE\cdot AC}{AF\cdot AD} \quad (6)$$Little claim is that $ECDF$ is cyclic; $\angle{CEF}=\angle{AEF}=\angle{PAC}=\angle{ADC}=180^\circ-\angle{CDF}$. $\square$ Thus, $$\frac{AC}{AD}=\frac{AF}{AE} \quad (7)$$By $(6)$ and $(7)$, we conclude that $$\frac{EB}{BF}=\frac{AE\cdot AC}{AF\cdot AD}=\frac{AE\cdot AF}{AF\cdot AE}=1.$$