Given a convex quadrilateral $ABCD$ with $AB = BC $and $\angle ABD = \angle BCD = 90$.Let point $E$ be the intersection of diagonals $AC$ and $BD$. Point $F$ lies on the side $AD$ such that $\frac{AF}{F D}=\frac{CE}{EA}$.. Circle $\omega$ with diameter $DF$ and the circumcircle of triangle $ABF$ intersect for the second time at point $K$. Point $L$ is the second intersection of $EF$ and $\omega$. Prove that the line $KL$ passes through the midpoint of $CE$. Proposed by Mahdi Etesamifard and Amir Parsa Hosseini - Iran
Problem
Source: Intermediate p3
Tags: Intermediate p3, geometry
BarisKoyuncu
30.12.2021 19:37
Let $M$ be the midpoint of $[EC]$ and let $AB\cap CD=S$. Let $\angle BCA=\angle BAC=\alpha$.
By Angle-Chasing, we get $\angle CBD=90^\circ-2\alpha, \angle DCE=90^\circ-\alpha, \angle DEC=\angle AEB=90^\circ-\alpha$. Hence, $|DC|=|DE|$. So, $\angle EDM=\angle CDM=\alpha$ and $DM\perp CE$. Also, $\angle ASC=90^\circ-2\alpha$.
By the Law of Sines in the triangles $BCE$ and $BAE$, we get $\frac{|CE|}{\sin{(90^\circ-2\alpha)}}=\frac{|BE|}{\sin \alpha}=\frac{|EA|}{\sin{90^\circ}}=|EA|\Rightarrow \frac{|AF|}{|FD|}=\frac{|CE|}{|EA|}=\sin{(90^\circ-2\alpha)}$.
$\frac{|AB|}{|BS|}=\frac{|BC|}{|BS|}=\frac{\sin{(90^\circ-2\alpha)}}{\sin{90^\circ}}=\sin{(90^\circ-2\alpha)}=\frac{|AF|}{|FD|}\Rightarrow BF//SD$.
Hence, $\angle FBD=\angle BDC=2\alpha\Rightarrow \angle ABF=90^\circ-2\alpha$.
By the Law of Sines in the triangles $AKF$ and $DKF$, we get
$$\frac{|KF|}{\sin{(KAF)}}=\frac{|AF|}{\sin{(AKF)}}=\frac{|AF|}{\sin{(ABF)}}=\frac{|AF|}{\sin{(90^\circ-2\alpha)}}\Rightarrow \frac{|KF|}{\sin{(KAF)}}=\frac{|AF|}{\sin{(90^\circ-2\alpha)}}$$$$\frac{|KF|}{\sin{(KDF)}}=\frac{|FD|}{\sin{(FKD)}}=\frac{|FD|}{\sin{90^\circ}}=|FD|\Rightarrow \frac{|KF|}{\sin{(KDF)}}=|FD|$$$$\Rightarrow \frac{|KF|}{\sin{(KAF)}}\cdot \frac{\sin{(KDF)}}{|KF|}=\frac{|AF|}{\sin{(90^\circ-2\alpha)}}\cdot \frac{1}{|FD|}=\frac{\sin{(KAF)}}{\sin{(KDF)}}=\frac{|AF|}{|FD|}\cdot \frac{1}{\sin{(90^\circ-2\alpha)}}$$$$=(90^\circ-2\alpha)\cdot \sin{(90^\circ-2\alpha)}=1\Rightarrow \sin{(KAF)}=\sin{(KDF)}\Rightarrow |KA|=|KD|$$
$\angle AKD=\angle AKF+\angle FKD=\angle ABF+\angle FKD=(90^\circ-2\alpha)+90^\circ=180^\circ-2\alpha\Rightarrow \angle KAD=\angle KDA=\alpha$.
$\angle FDE=\angle FDK+\angle KDE=\alpha+\angle KDE=\angle MDE+\angle KDE=\angle MDK\Rightarrow \angle FDE=\angle KDM$.
$\frac{|KD|}{|FD|}=\cos \alpha=\frac{|MD|}{|ED|}\Rightarrow \frac{|FD|}{|DE|}=\frac{|KD|}{|DM|}; \angle FDE=\angle KDM\Rightarrow \triangle FDE\sim \triangle KDM$.
Hence, $\angle MKD=\angle EFD=\angle LFD$.
$\angle MKD+\angle DKL=\angle LFD+\angle DKL=180^\circ (KLFD\text{ is cyclic})$. Thus; $K,L,M$ are collinear.
Hence, $KL$ bisects $[EC]$.
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wayneyam
20.05.2022 10:57
If we let $M$ be the foot from $F$ to $BD$, then it suffices to prove $AE$ is parellel to $KM$ (because of Reim's on $(DLE)$ and $(DLF)$). My solution after this is messy, can anyone help me finish the proof? I also got $BF \perp BC$ by phantom point, if that helps...
Tafi_ak
27.07.2022 12:12
wayneyam wrote: If we let $M$ be the foot from $F$ to $BD$, then it suffices to prove $AE$ is parellel to $KM$ (because of Reim's on $(DLE)$ and $(DLF)$). I got $BF \perp BC$ by phantom point, if that helps... Then it is sufficient to prove $\triangle ABC\sim\triangle DKA$. It is not easy using the given ratio directly to prove the similarity. So take a phantom point $K'$ such that $\triangle ABC\sim\triangle DK'A$ and $B,K'$ are the same side of $AD$. Now since $E,F$ are the corresponding point. So $\triangle ABE\sim\triangle DK'F, \triangle CBE\sim\triangle AK'F$. Then finish by angle chasing.
levifb
05.02.2023 22:52
Claim 1: $\angle FBC = 90 ^\circ$ and $DE = DC$. Proof: Law of Sines in $\triangle ABE$ and $\triangle BCE$ gives us $\dfrac{AF}{FD} = \dfrac{CE}{EA} = \dfrac{BC}{BA} \cdot \dfrac{sin~CBE}{sin~90^\circ} = sin~CBE$, and Law of Sines in $\triangle ABF$ and $\triangle BFD$ gives us $sin~CBD = sin~CBE = \dfrac{AF}{FD} = \dfrac{BA}{BD} \cdot tan~ABF \Rightarrow \dfrac{CD}{BD} = \dfrac{BA}{BD} \cdot tan~ABF \Rightarrow tan~ABF = \dfrac{CD}{BA} = \dfrac{CD}{BC} = tan~CBD \Rightarrow$ $ \angle ABF = \angle CBD \Rightarrow \angle FBC = 90^\circ$. If $Z = BF \cap AC$, we have $\angle BZC = \angle ECD = \angle BEZ = \angle DEC$ since $BZ // CD$ and $BE = BZ$ from $\triangle EBC \equiv \triangle ZBA$. So $DE = DC, ~~~Q.E.D.$ Claim 2: $\triangle ABC$ ~ $\triangle AKD$ Proof: Law of sines in $\triangle AKF$ and $\triangle AFD$ gives us $sin~CBE = \dfrac{AF}{FD} = \dfrac{AK}{DK} \cdot \dfrac{sin~AKF}{sin~90^\circ}$, but $\angle AKF = \angle ABF = \angle CBE$(Claim 1), so $AK = DK$. Note that $\angle AKD = 90^\circ + \angle AKF = 90^\circ + \angle ABF = \angle ABC$, so $\triangle ABC$ ~ $\triangle AKD~~~Q.E.D.$ Claim 3: if $BD \cap \omega = G$, $KG // AC$. Proof: Note that $\angle EGK = \angle KFD = \angle KAF + AKF = \angle BAC + \angle ABF$ and $\angle BEA = \angle CBE + \angle BCA \Rightarrow \angle EGK = \angle BEA \Rightarrow KG // AC ~~~ Q.E.D.$ Now, let $H$ be the second intersection of $\omega$ and the parallel line to $AC$ through $F$, $X$ the midpoint of $CE$, $X' = KL \cap DH$. With Pascal Theorem on $(HDGKLF)$, we get that $E, X'$ and $P_{\infty}(AC)$ are collinear, so $EX' // AC \Rightarrow X' = AC \cap DH = X$, since $\angle DHF = 90^\circ = \angle DXA \Rightarrow D,H,X$ are collinear. Hence, $KL$ bisects $CE$. $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\blacksquare$
GeorgeRP
10.10.2024 00:26
Let us denote with $V$ the foot from $F$ to $AB$ and with $H$ the one from $F$ to $BD$. Claim 1: $\angle{DVB}=\angle{AEB}$ Proof. It is sufficient to show that $\frac{BV}{BD}=\frac{BE}{AB}$. Let's denote $\phi=\angle{BDC}$. $\angle{CAB}=90-\frac{\angle{ABC}}{2}=\frac{\phi}{2}$. Then: $$DB=\frac{BC}{\sin\phi}=\frac{AB}{\sin\phi}$$$$BE=AB\tan{\frac{\phi}{2}}$$$$BV=FE=AB\frac{FD}{AD}=AB\frac{AE}{AC}=\frac{AB}{1+\frac{EC}{AC}}=\frac{AB}{1+\frac{\sin{(90-\phi)}}{\sin 90}\frac{BC}{AC}}=\frac{AB}{2\cos^2{\frac{\phi}{2}}}$$The desired directly follows as $$\frac{BV}{BD}=\frac{\sin\phi}{2\cos^2{\frac{\phi}{2}}}=\tan{\frac{\phi}{2}}=\frac{BE}{AB} \text{. } \square$$ By radical axes on $(ABKF),(DHKF),(ABD)$ we get that $FK, AB$ and the tangent at $D$ to $(FKD)$ meet at a point $R$. $\angle{FVR}=\angle{FDR}=90 \Rightarrow$ $VFDR$ is cyclic. Claim 2: $KH||AC$ Proof. $\angle{AEB}=\angle{DVR}=\angle{DFR}=\angle{KHE}\text{. } \square$ Define $M$ as the intersection of $LK$ with $EC$. Claim 3: $M$ is the midpoint of $EC$ Proof. As $\angle{KME}=\angle{HKM}=\angle{LDE}$ we get that $LEMD$ is cyclic. So $DM\perp EC$. But $\angle{DEC}=\angle{DCE}\Rightarrow DE=DC$, implying the desired, with which we are done with the problem. $\square$
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