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Source: Mathley

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doxuanlong15052000

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doxuanlong15052000

#1 h Jan 7, 2016, 5:05 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle. Points $E$ and $F$ are outside the triangle such that $EA=EC$ and $FA=FB$ and $\angle AEC=\angle AFB$ . Let $H$ and $K$ be the orthocenters of the triangles $AEC$ and $AFB$ respectively. Prove that $EF, HK, BC$ are concurrent.

This post has been edited 1 time. Last edited by doxuanlong15052000, Jul 10, 2016, 4:06 PM

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#2 h Jan 7, 2016, 1:43 PM • 1 Y

Y by Adventure10

Let $P =BK \cap CH,Q=BF \cap CE$ and $O$ is circumcenter of triangle $ABC$
By Desargue's Theorem on $\triangle{BFK} ,\triangle{CHE}$
We need to proof $P,O,Q$ collinear
Let $BO$ cut $PQ$ at $T$
We will proof that $\angle{TCB} =90-\angle{BAC}$
Let $\angle{CAE} =\angle{ACE} =\angle{BAF} =\angle{ABF} =m$
We get $\angle{BFA} =\angle{CAE} =180-2m$
Simple angle-chasing gives us $\angle{QBC} =180-m-\angle{B}$ and $\angle{QCB} =180-m-\angle{C}$
By Ceva's Theorem with cevian $PQ,BQ,CQ$ of triangle $PBC$
We get $\frac{sin(\angle{BPT})}{sin(\angle{CPT})} \frac{sin(270-m)}{sin(m+\angle{C})} \frac{sin(m+\angle{B})}{sin(270-m)} =1$
So $\frac{sin(\angle{BPT})}{sin(\angle{CPT})} =\frac{sin(m+\angle{C})}{sin(m+\angle{B})}$
By Ceva's Theorem with cevian $PT,BT,CT$ of triangle $PBC$
Let $\angle{BCT} =t$ give $\angle{ACT} =\angle{C} -t$
We get $\frac{sin(\angle{BPT})}{sin(\angle{CPT})} \frac{sin(90-m+\angle{C}-t)}{sin(t)} \frac{sin(90-\angle{A})}{sin(m+\angle{C})} =1$
Plugging value of $\frac{sin(\angle{BPT})}{sin(\angle{CPT})}$
Give us $\frac{sin(90-\angle{A})}{sin(m+\angle{B})} =\frac{sin(t)}{sin(90-m+\angle{C} -t)}$
We easily get that $t=90-\angle{A}$ is uniquely satisfy the equation.
So $\angle{TCB} =90-\angle{A}$
Which mean $T=O$

Another result
Let $H$ is orthocenter of triangle $ABC$
Proof that $P,Q,A,B,C,H$ lie on the same hyperbola.

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doxuanlong15052000

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doxuanlong15052000

#3 h Jan 7, 2016, 2:18 PM • 2 Y

Y by Adventure10, Mango247

Thank you !!

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#4 h Jan 7, 2016, 3:08 PM • 2 Y

Y by Adventure10, Mango247

Try to proof that $A,B,C,H,P,Q$ lie on the same hyperbola or
Locus of point $Q$ when we varied point $E$ on perpendicular bisector of $AC$ is hyperbola pass through $A,B,C,H$

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PROF65

#5 h Jan 7, 2016, 11:22 PM • 1 Y

Y by Adventure10

First the condition of $E$ and $F$ isn't sufficient . Either $BFK$ and $CEH$ overlap $ABC$ at the same time or they don't overlap it.
Let $S=BF \cap CE$ and $FK \cap EH=O,T =BK \cap OS,T' =CH \cap OS,U =BC \cap OS$ ( $O$ circumcenter of triangle $ABC$ ).
WLG consider $BFA$ and $ACH$ acute triangle , $\widehat{ABF}=\widehat{ACE}=\alpha ,\widehat{KBF}=\widehat{HCE}= \frac{\pi }{2}-\alpha$ .
$\frac{\sin \widehat{UBS}}{\sin \widehat{UBT}}:\frac{\sin \widehat{OBS}}{\sin \widehat{OBT}}= \frac{\sin (\hat B + \alpha) }{\sin (\hat B +\frac{\pi}{2}-\alpha)}:\frac{\sin (\frac{\pi}{2}-\hat C+\alpha)}{\sin (\frac{\pi}{2}-\hat C+ \frac{\pi}{2}-\alpha )} =\frac{\sin (\hat B + \alpha) }{\cos (\hat B -\alpha)}:\frac{\cos (\hat C-\alpha)}{\sin (\hat C+\alpha )}$
similarly we got $\frac{\sin \widehat{UCS}}{\sin \widehat{UCT'}}:\frac{\sin \widehat{OCS}}{\sin \widehat{OCT'}}=\frac{\sin (\hat B + \alpha) }{\cos (\hat B -\alpha)}:\frac{\cos (\hat C-\alpha)}{\sin (\hat C+\alpha )}$
which means $B(U,O;S,T)=C(U,O;S,T')$ then $T=T'$ finaly by desargues we conclude
WCP

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#7 h Jan 8, 2016, 1:22 AM • 4 Y

Y by doxuanlong15052000, enhanced, Adventure10, Mango247

Generalization : Given two inversely similar triangles $\triangle AB_1C_1$ and $\triangle AB_2C_2$ . Let $H_1,$ $H_2$ be the orthocenter of $\triangle AB_1C_1,$ $\triangle AB_2C_2$ , respectively. Then $B_1B_2,$ $C_1C_2,$ $H_1H_2$ are concurrent.

Proof : Let $P$ be the point such that $PB_1$ $=$ $PB_2,$ $PC_1$ $=$ $PC_2$ and let $T$ $\equiv$ $B_1B_2$ $\cap$ $C_1C_2$ . From 2000 IMO Shortlist G6 $\Longrightarrow$ $\measuredangle C_1PC_2$ $=$ $2 \measuredangle C_1B_1A$ , so $\measuredangle TC_1P$ $=$ $90^{\circ}$ $-$ $\measuredangle C_1B_1A$ $=$ $\measuredangle H_1C_1B_1$ . Analogously, we can prove $\measuredangle TB_1P$ $=$ $\measuredangle H_1B_1C_1$ , so $H_1$ and $P$ are isogonal conjugate WRT $\triangle TB_1C_1$ $\Longrightarrow$ $H_1$ lies on the isogonal conjugate $\tau$ of $TP$ WRT $\angle (B_1B_2,C_1C_2)$ . Similarly, we can prove $H_2$ lies on $\tau$ , so $T,$ $H_1,$ $H_2$ are collinear. i.e. $B_1B_2,$ $C_1C_2,$ $H_1H_2$ are concurrent

Remark : See also here (another generalization) and here (generalization of Generalization).

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