Hi, my name is dr. Ilham Kosasih, I am from Indonesia, if i have a questions about Math, can i ask u?? , would u pliz help me?? btw, my email is masterpool4ever@yahoo.com, i love math.

Seems that M.T.

Hello! a) $ Q\in EF$, $ \left\{P_{1}\right\}=CP\cap AB$ and $ \left\{P_{2}\right\}=DP\cap AB$. $ \frac{P_{1}A}{P_{1}B}=\frac{CA\cdot sin\left(\angle ACP_{1}\right)}{CB\cdot sin\left(\angle BCP_{1}\right)}=\frac{CA}{CB}\cdot\frac{sin\left(\angle DCF\right)}{sin\left(\angle QCF\right)}=$ $ =\frac{CA}{CB}\cdot\frac{DF\cdot sin\left(\angle CDF\right)}{QF\cdot sin\left(\angle CQF\right)}=\frac{CA}{CB}\cdot\frac{DF}{QF}\cdot\frac{sin\left(\angle CEF\right)}{sin\left(\angle CQE\right)}=$ $ =\frac{CA}{CB}\cdot\frac{DF}{QF}\cdot\frac{CQ}{CE}$ (1) $ \frac{P_{2}A}{P_{2}B}=\frac{DA\cdot sin\left(\angle ADP_{2}\right)}{DB\cdot sin\left(\angle BDP_{2}\right)}=\frac{DA}{DB}\cdot\frac{sin\left(\angle QDE\right)}{sin\left(\angle CDE\right)}=$ $ =\frac{DA}{DB}\cdot\frac{QE\cdot sin\left(\angle DQE\right)}{CE\cdot sin\left(\angle DCE\right)}=\frac{DA}{DB}\cdot\frac{QE}{CE}\cdot\frac{sin\left(\angle DQF\right)}{sin\left(\angle DFE\right)}=$ $ =\frac{DA}{DB}\cdot\frac{QE}{CE}\cdot\frac{DF}{DQ}$ (2) (1), (2) $ \Rightarrow\frac{P_{1}A}{P_{1}B}=\frac{P_{2}A}{P_{2}B}\Leftrightarrow \frac{CA}{CB}\cdot\frac{DF}{QF}\cdot\frac{CQ}{CE}=\frac{DA}{DB}\cdot\frac{QE}{CE}\cdot\frac{DF}{DQ}\Leftrightarrow$ $ \Leftrightarrow\frac{QC}{QE}\cdot\frac{QD}{QF}=\frac{CB}{DB}\cdot\frac{DA}{CA}\Leftrightarrow$ $ \Leftrightarrow\frac{sin\left(\angle QEC\right)}{sin\left(\angle QCE\right)}\cdot\frac{sin\left(\angle QFD\right)}{sin\left(\angle QDF\right)}=\frac{sin\left(\angle CDB\right)}{sin\left(\angle DCB\right)}\cdot\frac{sin\left(\angle DCA\right)}{sin\left(\angle CDA\right)}\Leftrightarrow$ $ \Leftrightarrow\frac{sin\left(\angle QCD\right)}{sin\left(\angle QCE\right)}\cdot\frac{sin\left(\angle QDC\right)}{sin\left(\angle QDF\right)}=1$. The last relation is true. So $ P_{1}=P_{2}=P\Rightarrow P\in AB$. b) $ P\in AB$, $ \left\{Q_{1}\right\}=BC\cap EF$ and $ \left\{Q_{2}\right\}=AD\cap EF$. $ \frac{Q_{1}E}{Q_{1}F}=\frac{\frac{Q_{1}C\cdot sin\left(\angle Q_{1}CE\right)}{sin\left(\angle Q_{1}EC\right)}}{\frac{Q_{1}C\cdot sin\left(\angle Q_{1}CF\right)}{sin\left(\angle Q_{1}FC\right)}}=\frac{sin\left(\angle Q_{1}CE\right)}{sin\left(\angle Q_{1}EC\right)}\cdot\frac{sin\left(\angle Q_{1}FC\right)}{sin\left(\angle Q_{1}CF\right)}=$ $ =\frac{sin\left(\angle Q_{1}CD\right)}{sin\left(\angle CDB\right)}\cdot\frac{sin\left(\angle CDE\right)}{sin\left(\angle BCP\right)}=\frac{sin\left(\angle BCD\right)}{sin\left(\angle CDB\right)}\cdot\frac{sin\left(\angle BDP\right)}{sin\left(\angle BCP\right)}=$ $ =\frac{BD}{BC}\cdot\frac{sin\left(\angle BDP\right)}{sin\left(\angle BCP\right)}=\frac{BP\cdot sin\left(\angle BPD\right)}{BP\cdot sin\left(\angle BPC\right)}=\frac{sin\left(\angle BPD\right)}{sin\left(\angle BPC\right)}$ (3) $ \frac{Q_{2}E}{Q_{2}F}=\frac{\frac{Q_{2}D\cdot sin\left(\angle Q_{2}DE\right)}{sin\left(\angle Q_{2}ED\right)}}{\frac{Q_{2}D\cdot sin\left(\angle Q_{2}DF\right)}{sin\left(\angle Q_{2}FD\right)}}=\frac{sin\left(\angle Q_{2}DE\right)}{sin\left(\angle Q_{2}ED\right)}\cdot\frac{sin\left(\angle Q_{2}FD\right)}{sin\left(\angle Q_{2}DF\right)}=$ $ =\frac{sin\left(\angle ADP\right)}{sin\left(\angle DCF\right)}\cdot\frac{sin\left(\angle DCA\right)}{sin\left(\angle Q_{2}DC\right)}=\frac{sin\left(\angle ADP\right)}{sin\left(\angle ACP\right)}\cdot\frac{sin\left(\angle DCA\right)}{sin\left(\angle CDA\right)}=$ $ =\frac{sin\left(\angle ADP\right)}{sin\left(\angle ACP\right)}\cdot\frac{AD}{AC}=\frac{AP\cdot sin\left(\angle APD\right)}{AP\cdot sin\left(\angle APC\right)}=\frac{sin\left(\angle APD\right)}{sin\left(\angle APC\right)}$ (4) (3), (4) $ \Rightarrow\frac{Q_{1}E}{Q_{1}F}=\frac{Q_{2}E}{Q_{2}F}\Leftrightarrow\frac{sin\left(\angle BPD\right)}{sin\left(\angle BPC\right)}=\frac{sin\left(\angle APD\right)}{sin\left(\angle APC\right)}$. The last relation is true. So $ Q_{1}=Q_{2}=Q\Rightarrow Q\in EF$. Best regards, Petrisor Neagoe

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Petry wrote: b) $ P\in AB$, $ \left\{Q_{1}\right\}=BC\cap EF$ and $ \left\{Q_{2}\right\}=AD\cap EF$. Nice proof! Just wanted to point out that once you've proven (a), you can just prove that $P\in AB$ must be unique (for instance, by noting $\angle{PDC}$ and $\angle{PCD}$ increase with $EF$), so no need to go through all the law of sines again for (b). --- Here's another solution. Let $T=DB\cap CA$ so that $Q$ is the $T$-excenter of $\triangle{TDC}$. The convexity of $ABCD$ yields $T>C,D$. We are given that $P$ exists, so because $DA,CB$ are external bisectors of $\angle{PDE},\angle{PCF}$, $Q,T$ must lie on opposite sides of $DE$ and $CF$. Let $\theta = \angle{TDE}$. Since $DCEF$ is cyclic, $\angle{PDC}=\angle{TDE}=\theta=\angle{TCF}=\angle{PCD}$. Hence $P$ varies (with $\theta<90^\circ$) along the perpendicular bisector $\ell$ of $DC$, so $P\in AB$ for exactly one value of $\theta$ (say $\alpha$). Similarly, we always have $\angle{TFE}=C$ and $\angle{TEF}=D$, so $Q\in EF$ for a unique value of $\theta$ (say $\beta$). It clearly suffices to show that $\alpha=\beta$. First suppose $\theta=\alpha$, so $FE=r_T(\csc{C}+\csc{D}) = 2[TCD](\csc{C}+\csc{D})/(c+d-t)=t(d+c)/(c+d-t)$ (by the sine area formula). Then \[ \frac{FE}{\sin\theta} = \frac{FE}{\sin{FDE}}=\frac{CD}{\sin{CFD}} = \frac{t}{\sin(T+\theta)} \implies \frac{\sin T}{\tan\alpha}+\cos{T} = \frac{c+d-t}{c+d}, \]where we use $\sin(T+\theta)=\sin{T}\cos\theta+\cos{T}\sin\theta$. Now suppose $\theta=\beta$, and let $A',P',B'$ be the feet from $A,P,B$ to $DC$. Then \[ \frac{t}{2}\tan\beta = DP'\tan\beta = PP' = \frac{2[PA'B']}{A'B'} = \frac{2[P'AB]}{A'B'} = \frac{[DAB]+[CAB]}{TA\cos{C}+TB\cos{D}}. \]By the angle bisector theorem, $\frac{TA}{TC}=\frac{TA}{CA-TA}=\frac{TD}{CD-TD}$, so $TA = \frac{dc}{t-c}$ and $CA = TA+d=\frac{td}{t-c}$. Similarly, $TB = \frac{cd}{t-d}$ and $DB = \frac{tc}{t-d}$, so \begin{align*} \tan\beta = \frac{2}{t}\frac{[DAB]+[CAB]}{TA\cos{C}+TB\cos{D}} &= \frac{(t-c)(t-d)\sin{T}}{tcd}\frac{DB\cdot TA+CA\cdot TB}{TA\cos{C}+TB\cos{D}} \\ &= \frac{(t-c)(t-d)\sin{T}}{tcd}\frac{\frac{tcd}{(t-c)(t-d)}(c+d)}{(t-d)\cos{C}+(t-c)\cos{D}} \\ &= \frac{(c+d)\sin{T}}{t(\cos{C}+\cos{D}) - t}. \end{align*}Thus \begin{align*} \frac{\sin{T}}{\tan\beta}+\frac{t}{c+d} &= \frac{\sin{T}\cos{C}+\sin{T}\cos{D}}{\sin{C}+\sin{D}} \\ &= \frac{\sin(T+C)-\cos{T}\sin{C}+\sin(T+D)-\cos{T}\sin{D}}{\sin{C}+\sin{D}} = 1-\cos{T} = \frac{\sin{T}}{\tan\alpha}+\frac{t}{c+d}, \end{align*}so we're done (as $\theta$ must be acute in order for $P$ to exist).

This can be done using harmonic range of points. And i wonder if the condition "$CEFD$ is concyclic" can be deleted?

P-H-David-Clarence wrote: This can be done using harmonic range of points. And i wonder if the condition "$CEFD$ is concyclic" can be deleted? You are right, the condition "$CEFD$ is cyclic" can be deleted. Here's a short proof using barycentric coordinates. Let $AC \cap BD = X$, and $X=(1,0,0), C=(0,1,0), D=(0,0,1)$, $a=CD, b=DX, c=XD$ $Q$ is $X$- excenter of triangle $XCD$, so $Q=(-a:b:c)$ $E$ is on $XC$, so let $E=(t, 1-t, 0)$ for some real number $t$, and similary let $F=(u, 0, 1-u)$ for some real number $u$ $P' = CF \cap DE$, so $P'=(ut:u(1-t):t(1-u))$ $P$ is isogonal conjugate of $P'$, so $P=(a^2 : \frac{t}{1-t} b^2 : \frac{u}{1-u} c^2)$. ($u, t$ are not $0, 1$) $A = DQ \cap XC$, so $A=(-a:b:0)$, and similary $B=(-a:0:c)$ $P$ is on $AB \iff \det\begin{pmatrix} -a & b & 0 \\ -a & 0 & c \\ a^2 & \frac{t}{1-t} b^2 & \frac{u}{1-u} c^2 \end{pmatrix}=0 \iff a+ b \frac{t}{1-t} + c \frac{u}{1-u} =0$ $Q$ is on $EF $ $\iff \det \begin{pmatrix} -a & b & c \\ t & 1-t & 0 \\ u & 0 & 1-u \end{pmatrix} =0 \iff -a(1-t)(1-u)-bt(1-u)-cu(1-t)=0 \iff a+ b \frac{t}{1-t} + c \frac{u}{1-u} =0$ So, $P$ is on $AB$ iff $Q$ is on $EF$, whether $CDFE$ is concyclic or not.