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Source: Romanian JBTST VI 2007, problem 3

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#1 h Jun 8, 2007, 6:59 AM • 7 Y

Y by narutomath96, Adventure10, Mango247, and 4 other users

Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$ . Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$ . Prove that $AF, DE$ and $BC$ are concurrent.

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#2 h Jun 8, 2007, 7:30 PM • 8 Y

Y by BobaFett101, Adventure10, Shinichi-123, Mango247, and 4 other users

Denote the points $X \in AE \cap BD$ , $Y \in AE \cap BC$ , $Z \in AE \cap DF$ and $T \in DF \cap BC$ .

In $\triangle{AEC}$ , we observe that the lines $AF$ , $DE$ and $BC$ are concurrent, if and only if, the division $(AYEZ)$ is harmonic.

Since the quadrilateral $XYTD$ is cyclic, $\tan{XYB}= \tan{XDZ}$ , which is equivalent to
$\frac{XB}{XY}=\frac{XZ}{XD}\Longleftrightarrow XB \cdot XD = XY \cdot XZ.$
Due to the similarity of the triangles $\triangle{XAB}$ and $\triangle{XDA}$ , we have that $XA^{2}=XB \cdot XD$ , so $XA^{2}=XY \cdot XZ$ , which by using $XA=XF$ , it is equivalent with $\frac{YA}{YE}=\frac{ZA}{ZE}$ , i.e. the division $(AYEZ)$ is harmonic.

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#3 h Jun 9, 2007, 7:43 PM • 6 Y

Y by Adventure10, Mango247, and 4 other users

We denote as $K,$ $L,$ the intersection points of $BC,$ from $DF,$ $AE$ respectively and also the points $X\equiv AC\cap BE,$ $Y\equiv AB\cap CE,$ $Z\equiv AE\cap XY.$

From the complete quadrilateral $ABECXY,$ we conclude that the points $L,$ $Z,$ are harmonic conjugates, with respect to the points $A,$ $E.$

Because of $AD\perp AB,$ $DE\perp BE,$ $DK\perp KB,$ we conclude that the pentagon $ABEKD$ is cyclic ( taken as diameter the segment $BD$ ) and so, we have that the segment line $KB,$ is the angle bisector of the angle $\angle AKE,$ because of $AB = BE.$

So, because of $DK\perp KB,$ we conclude that the segment line $DF,$ passes through the point $Z,$ as the external angle bisector of the triangle $\bigtriangleup KAE,$ through vertex $K.$

Because of now, the points $X,$ $Y,$ $Z$ are collinear, based on the Desarques’s theorem, we conclude that the triangles $\bigtriangleup BEA,$ $\bigtriangleup CDF,$ are perspective $($ $X\equiv BE%Error. "capCD" is a bad command. ,$ $Y\equiv BA\cap CF,$ $Z\equiv AE\cap DF$ $).$

Hence, the segment lines $AF,$ $DE,$ $BC$ are concurrent at one point and the proof is completed.

Kostas Vittas.

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#4 h Jun 18, 2007, 9:27 AM • 7 Y

Y by Adventure10, Mango247, and 5 other users

Here is another solution to this nice problem:
Let $DF\cap BC= O$ , $AE\cap BD=M$ , $AE\cap BC=N$ and assume $AD\geq DC$ .
Let $\measuredangle{CBD}=x$ , $\measuredangle{DBA}=y$ , and $\measuredangle{BCA}=c$ . It`s clear that $x+y+c=90^{\circ}$
Because triangle $BAM$ is right-angled, we get $\measuredangle{BAM}=c+x$ .
Because the quadrilateral $ABDO$ is cyclic, we have $\measuredangle{OAD}=\measuredangle{OBD}=x$ ,
thus $\measuredangle{OAE}=y-x$ .
In isosceles triangle $ABE$ , we have $\measuredangle{BAE}=c+x$ , therefore $\measuredangle{ABE}=2y$ .
But $\measuredangle{ABN}=x+y$ , therefore $\measuredangle{EBO}=y-x=\measuredangle{OAE}$ .
This means that quadrilaterel $OABE$ is cyclic, and thus the pentagon $ADOEB$ is cyclic.
This means that $\measuredangle{EDB}=\measuredangle{EAB}=c+x\Longrightarrow \measuredangle{ADE}=2(c+x)\Longrightarrow \measuredangle{EDF}=y-x$ .
In triangle $ABE$ , $\ds N\in (AE)\Longrightarrow \frac{AN}{NE}=\frac{AB}{BE}\cdot\frac{\sin \measuredangle{ABN}}{\sin\measuredangle{EBN}}\Longrightarrow \frac{AN}{NE}=\frac{\sin(x+y)}{\sin(y-x)}$
In triangle $EDC$ , $\ds F\in (EC)\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin\measuredangle{EDF}}{\sin\measuredangle{FDC}}\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin(y-x)}{\sin(x+y)}$
By multiplying the last two relations we obtain $\ds \frac{AN}{NE}\cdot\frac{EF}{FC}\cdot\frac{CD}{DA}=\frac{ED}{DA}=1$,
therefore, by the converse of Ceva`s theorem, the lines $AF, CN, ED$ are concurent, QED.

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Sergey

#5 h May 4, 2009, 6:27 PM • 2 Y

Y by Adventure10, Mango247

pohoatza wrote:

Due to the similarity of the triangles $\triangle{XAB}$ and $\triangle{XDA}$ , we have that $XA^{2} = XB \cdot XD$ , so $XA^{2} = XY \cdot XZ$ , which by using $XA = XF$ , it is equivalent with $\frac {YA}{YE} = \frac {ZA}{ZE}$ , i.e. the division $(AYEZ)$ is harmonic.

Please could you tell how exactly using $XA=XF$ we get $\frac {YA}{YE} = \frac {ZA}{ZE}?$

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#6 h May 6, 2009, 4:17 PM • 3 Y

Y by Adventure10, Mango247, XX-math-XX

Remark (a short commentary). It is a very nice application of the following remarkable harmonical division (see the nice Pohoatza's proof !) :

Quote:

Lemma. Let $w(O)$ , $w_o$ be two secant circles in $\{A,B\}$ so that $O\in w_o$ . For $P\in (AB)$ (segment !) denote $\{M,N,R\}$

such that $\{M,N\}\subset w$ , $R\in w_o$ and $N$ separates $P$ , $R$ . Then the division $\{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically.

Particular case. Let $ABC$ be a triangle with the orthocenter $H$ . Denote $D\in BC\cap AH$ and the intersections $N$ , $S$

between the line $AH$ and the circle with the diameter $[BC]$ . Then the division $\{\ A\ ,\ H\ ,\ N\ ,\ S\ \}$ is harmonically.

Proof. $PO\cdot PR = PA\cdot PB = PM\cdot PN\ \implies\ PO\cdot PR = PM\cdot PN\ \Longleftrightarrow$ the division $\{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically.

The following interesting problem is a nice consequence of the harmonical division which was mentioned in the above lemma :

Quote:

Let $w(O)$ , $w_o$ be two secant circles so that $O\in w_o$ . Consider $A\in w\cap w_o$ . For $P\in (AB)$ (segment !) denote $\{M,N,R\}$ such that

$\{M,N\}\subset w$ , $R\in w_o$ and $N$ separates $P$ , $R$ . For $C\in AB$ (line !) denote $\left\|\begin{array}{c} X\in MC\cap RA \\ \ Y\in NC\cap RA\end{array}\right\|$ .Then $MY\cap NX\cap CP\ne\emptyset$ .

Proof. From the upper lemma obtain that the division $\{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically, i.e. $\frac {\overline {PM}}{\overline {PN}} = - \frac {\overline {RM}}{\overline{RN}}$ . Apply the Menelaus' theorem

to the transversal $\overline {RXY}$ and the triangle $CMN\ : \ \frac {\overline{RM}}{\overline{RN}}\cdot \frac {\overline{YN}}{\overline{YC}}\cdot \frac {\overline{XC}}{\overline{XM}} = + 1$ $\implies$ $\frac {\overline{PM}}{\overline{PN}}\cdot \frac {\overline{YN}}{\overline{YC}}\cdot \frac {\overline{XC}}{\overline{XM}} = - 1$ , i.e. $MY\cap NX\cap CP\ne\emptyset$ .

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#7 h May 7, 2009, 6:27 AM • 2 Y

Y by Adventure10, Mango247

Sergey wrote:

pohoatza wrote:

Due to the similarity of the triangles $\triangle{XAB}$ and $\triangle{XDA}$ , we have that $XA^{2} = XB \cdot XD$ , so $XA^{2} = XY \cdot XZ$ , which by using $XA = XF$ , it is equivalent with $\frac {YA}{YE} = \frac {ZA}{ZE}$ , i.e. the division $(AYEZ)$ is harmonic.

Please could you tell how exactly using $XA = XF$ we get $\frac {YA}{YE} = \frac {ZA}{ZE}?$

There is a typo. He means $XA = XE,$ instead of $XA = XF.$

Kostas Vittas.

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Petry

#8 h May 7, 2009, 5:24 PM • 1 Y

Y by Adventure10

Hello!

$\left\{M\right\}=BE\cap DF$ , $\left\{N\right\}=BC\cap DF$ , $\left\{H\right\}=DE\cap BN$ ,
$\left\{S\right\}=MH\cap BD$ , $\left\{T\right\}=MH\cap AC$ .

$DE\bot MB$ and $BN\bot MD\Rightarrow$ the point $H$ is the orthocenter of the triangle $\Delta MBD\Rightarrow$
$\Rightarrow MS\bot BD$ and $\angle MSN=\angle MSE$ (1)
$\Delta ABS\equiv\Delta EBS\Rightarrow\angle ASB=\angle ESB\Rightarrow\angle AST=\angle MSE$ (2)
(1), (2) $\Rightarrow\angle MSN=\angle AST\Rightarrow$ the points $A,S,N$ are collinear.

Let's consider the triangles $\Delta SDA$ and $\Delta MEF$ .
$\left\{B\right\}=SD\cap ME$ , $\left\{C\right\}=DA\cap EF$ , $\left\{N\right\}=AS\cap FM$ and the points $B,C,N$ are collinear.
Based on the Desargues's theorem, we conclude that the triangles $\Delta SDA$ and $\Delta MEF$ are
perspective (the lines $SM,DE$ and $AF$ are concurrent at the point $H$ ).
Hence, the lines $AF,DE$ and $BC$ are concurrent at the point $H$ .

Best regards, Petrisor Neagoe

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#9 h May 7, 2009, 5:31 PM • 1 Y

Y by Adventure10

PhilAndrew wrote:

Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$ . Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$ . Prove that $AF, DE$ and $BC$ are concurrent.

Here is another approach:
Let $T\equiv DF \cap AE$ , $S\equiv AE\cap BC$ , $K\equiv AE\cap BD$ , $H\equiv DT\cap BC$ . In order to prove $AF,DE,BC$ are concurrent, we need to prove $(A,E,S,T) = - 1$ . Define $(BD)$ is the circle with diameter $BD$ , $(DS)$ is the circle with diameter $DS$ . It is easy to notice that $\{A,B,D,E,H\}\in (BD)$ and $\{K,D,H,S\}\in (DS)\Longrightarrow DH$ is the radical axis wrt $(BD),(DS)$ . Thus, $\mathcal {P}_{T/(BD)} = \mathcal {P}_{T/(DS)}$
$\Longrightarrow \overline{TS}.\overline {TK} = \overline {TE}.\overline {TA}$ , but $K$ is the midpoint of $AE$ , hence $(AEST) = - 1$ , which implies to the result of the problem.
Our proof is completed $\blacksquare$ .

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#10 h Dec 1, 2013, 11:56 PM • 2 Y

Y by Adventure10, Mango247

andyciup wrote:

Here is another solution to this nice problem:
Let $DF\cap BC= O$ , $AE\cap BD=M$ , $AE\cap BC=N$ and assume $AD\geq DC$ .
Let $\measuredangle{CBD}=x$ , $\measuredangle{DBA}=y$ , and $\measuredangle{BCA}=c$ . It`s clear that $x+y+c=90^{\circ}$
Because triangle $BAM$ is right-angled, we get $\measuredangle{BAM}=c+x$ .
Because the quadrilateral $ABDO$ is cyclic, we have $\measuredangle{OAD}=\measuredangle{OBD}=x$ ,
thus $\measuredangle{OAE}=y-x$ .
In isosceles triangle $ABE$ , we have $\measuredangle{BAE}=c+x$ , therefore $\measuredangle{ABE}=2y$ .
But $\measuredangle{ABN}=x+y$ , therefore $\measuredangle{EBO}=y-x=\measuredangle{OAE}$ .
This means that quadrilaterel $OABE$ is cyclic, and thus the pentagon $ADOEB$ is cyclic.
This means that $\measuredangle{EDB}=\measuredangle{EAB}=c+x\Longrightarrow \measuredangle{ADE}=2(c+x)\Longrightarrow \measuredangle{EDF}=y-x$ .
In triangle $ABE$ , $\ds N\in (AE)\Longrightarrow \frac{AN}{NE}=\frac{AB}{BE}\cdot\frac{\sin \measuredangle{ABN}}{\sin\measuredangle{EBN}}\Longrightarrow \frac{AN}{NE}=\frac{\sin(x+y)}{\sin(y-x)}$
In triangle $EDC$ , $\ds F\in (EC)\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin\measuredangle{EDF}}{\sin\measuredangle{FDC}}\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin(y-x)}{\sin(x+y)}$
By multiplying the last two relations we obtain $\ds \frac{AN}{NE}\cdot\frac{EF}{FC}\cdot\frac{CD}{DA}=\frac{ED}{DA}=1$,
therefore, by the converse of Ceva`s theorem, the lines $AF, CN, ED$ are concurent, QED.

Darn it this was exactly the same proof I just rediscovered a few minutes ago, which I originally wrote down a while ago but threw away the paper.

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#11 h Sep 9, 2014, 5:15 AM • 1 Y

Y by Adventure10

PhilAndrew wrote:

Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$ . Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$ . Prove that $AF, DE$ and $BC$ are concurrent.

The inspiration for this solution comes from Cosmin (pohoatza) in https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCAQFjAA&url=http%3A%2F%2Fdiendantoanhoc.net%2Fforum%2Findex.php%3Fapp%3Dcore%26module%3Dattach%26section%3Dattach%26attach_id%3D16951&ei=dokOVOH4E8S0yASt5YLgBQ&usg=AFQjCNHsxpkOva9b0YPm1Io5LkgocHpOCg&bvm=bv.74649129,d.aWw (Harmonic Divisions and its Applications).

Solution

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HadjBrahim-Abderrahim

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HadjBrahim-Abderrahim

#12 h Aug 19, 2016, 5:51 PM • 2 Y

Y by Adventure10, Mango247

PhilAndrew wrote:

Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$ . Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$ . Prove that $AF, DE$ and $BC$ are concurrent.

My solution. Let $X$ , $Y$ , and $Z$ be the intersections of the line $AE$ with the lines $BD$ , $BC$ , and $DF$ , respectively, and let $W$ be the intersection of the lines $DF$ and $BC.$ Because $E$ is the reflection of $A$ in the line $BD$ we have, $\angle DEB=\angle BAD=\angle DWB = 90^\circ$ , which means that the pentagon $ADWEB$ is Cyclic. Thus, we have $\angle YWE=\angle BWE= \angle BAE=\angle BAX=90^\circ - \angle XAD= \angle ADB=\angle AWB=\angle AWY$ and we know that the lines $YW$ and $WZ$ are perpendicular. Therefore, the cross ratio $(A,E,Y,Z)$ is harmonic bundle, which means that the lines $CB$ , $DE$ , and $AF$ are concurrent.

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#13 h Aug 20, 2016, 8:36 AM • 1 Y

Y by Adventure10

Sergey wrote:

pohoatza wrote:

Due to the similarity of the triangles $\triangle{XAB}$ and $\triangle{XDA}$ , we have that $XA^{2} = XB \cdot XD$ , so $XA^{2} = XY \cdot XZ$ , which by using $XA = XF$ , it is equivalent with $\frac {YA}{YE} = \frac {ZA}{ZE}$ , i.e. the division $(AYEZ)$ is harmonic.

Please could you tell how exactly using $XA=XF$ we get $\frac {YA}{YE} = \frac {ZA}{ZE}?$

it is newton theorem

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#14 h Apr 22, 2018, 4:40 PM • 2 Y

Y by Adventure10, Mango247

Let $DF \cap BC = K$
It is clear that $ADBK$ and $ADBE$ are cyclic which forces $AKEB$ also to be cyclic.
Now ,since $\angle EAB = \angle AEB ,BK$ bisects $\angle AKE$ .
Now,by blanchet's theorem in $\triangle CDF$ , $AF,DE,BC$ concur. $\blacksquare$

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#15 h Feb 8, 2020, 3:32 PM • 5 Y

Y by mueller.25, Dr_Vex, myh2910, sttsmet, Adventure10

Ig there was a need for a different solution. So Bumped.

PhilAndrew wrote:

Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$ . Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$ . Prove that $AF, DE$ and $BC$ are concurrent.

Let $DF\cap BC=T$ . So, $\{A,B,E,T,D\}$ all lie on a circle. Now we will use Ratio Lemma and Ceva's Theorem.
$\frac{AK}{KE}=\frac{AB\sin\angle BAC}{BE\sin\angle EBC}$ and $\frac{EF}{FC}=\frac{ED\sin\angle EDF}{CD\sin\angle FDC}$ . Now $\angle EDF=\angle EBC$ and $\angle ABC=\angle FDC$ .

So now applying Ceva's Theorem we get that

$\frac{AD}{DC}\cdot\frac{CF}{FE}\cdot\frac{EK}{KA}=\frac{AD}{DC}\cdot\frac{DC\sin\angle FDC}{ED\sin\angle EDF}\cdot\frac{BE\sin\angle EBC}{AB\sin\angle ABC}=\frac{AD}{DC}\cdot\frac{DC}{ED}\cdot{BE}{AB}=1\implies AF,DE,BC\text{ are concurrent.}\blacksquare$

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jrsbr

#17 h Feb 11, 2023, 3:38 PM

Y by

Let $P=AF\cap BC$ , $Q=DF\cap BC$ and $H=AE\cap BC$ . See that $Q\in (ABDE)$ and therefore
$(Q,H,C,P)\overset{A}{=}(G,E,C,F)\overset{Q}{=}(A,E,B,D)=-1.$ Furthermore, if $P'=DE\cap BC$ and $I=DH\cap (ABD)$
$(Q,H,C,P')\overset{D}{=}(Q,I,A,E)\overset{H}{=}(B,D,E,A)=-1,$ and thus $P=P'$ , as desired. $\blacksquare$

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#18 h Dec 21, 2023, 8:57 PM

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Here's a trig-ceva solution:

Suppose $\angle ABD = \alpha$ , and $\angle DBC = \beta$ . Also, let $\angle ECD = \theta_1$ , and $\angle ECB = \theta_2$ .
Now, trig-ceva on $\displaystyle \triangle BDC$ for the point $E$ gives $\frac{\sin \theta_1}{\sin \theta_2} = \frac{\sin(2 \alpha)}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin(\beta - \alpha)} \ \ \ \ \ -(1)$
Let $\angle FBC = x$ . Trig-Ceva on $\displaystyle \triangle BDC$ for the point $F$ gives $\frac{\sin \theta_1}{\sin \theta_2} \cdot \frac{\sin(x)}{\sin(\beta -x)} \cdot \frac{\sin(\alpha+ \beta)}{\cos \beta} = 1 \implies \frac{\sin(\beta - x)}{\sin(x)} = \frac{\sin \theta_1}{\sin \theta_2} \cdot \frac{\sin(\alpha+ \beta)}{\cos \beta} = \frac{\sin(2 \alpha)}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin(\beta - \alpha)} \cdot \frac{\sin(\alpha+ \beta)}{\cos \beta}$ where we used $(1)$ in the last step. Thus, we get
$\sin \beta \cdot \cot(x) = \cos \beta + \frac{2 \cdot \sin ^2 \alpha \cdot \cos \beta}{\sin (\alpha+\beta) \cdot \sin (\beta - \alpha)} \ \ \ \ \ -(2)$
Let $\angle FAC = y$ . Now, trig-ceva on $\displaystyle \triangle ABC$ for the point $F$ gives $\frac{\sin y}{\cos y} = \frac{\sin \theta_1}{\sin \theta_2} \cdot \frac{\sin x}{\sin(\alpha + \beta -x)} = \frac{\sin(2 \alpha)}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin(\beta - \alpha)} \cdot \frac{\sin x}{\sin(\alpha + \beta -x)} \ \ \ \ \ -(3)$ where we used $(1)$ in the last equality.

Thus, finally we have by Trig Ceva in $\displaystyle \triangle ABD$ for the lines $AF, DE, BC$ , that $AF,DE,BC$ are concurrent if and only if
$\frac{\sin y}{\cos y} \cdot \frac{\sin(\alpha+\beta)}{\sin \beta} \cdot \frac{\cos \alpha}{\sin(2 \alpha)} = 1 \iff \frac{\sin(2 \alpha)}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin(\beta - \alpha)} \cdot \frac{\sin x}{\sin(\alpha + \beta -x)} \cdot \frac{\sin(\alpha+\beta)}{\sin \beta} \cdot \frac{\cos \alpha}{\sin(2 \alpha)} = 1$ Cancelling out terms, and taking $\sin \beta$ and terms involving $x$ to the other side, we are reduced to showing that
$\sin \beta \cdot\sin(\alpha + \beta) \cdot \cot x - \sin \beta \cdot \cos(\alpha + \beta) = \frac{\sin \alpha \cdot \sin(\alpha+\beta)}{\sin(\beta - \alpha)}$ Plugging in the value of $\sin \beta \cdot \cot x$ from $(2)$ , we get $LHS = \frac{2 \cdot \sin ^2 \alpha \cdot \cos \beta}{\sin (\beta - \alpha)} + \sin(\alpha + \beta) \cdot \cos \beta - \sin \beta \cdot \cos(\alpha + \beta) = \frac{2 \cdot \sin ^2 \alpha \cdot \cos \beta}{\sin (\beta - \alpha)} + \sin \alpha = \frac{\sin \alpha \cdot \sin(\alpha+\beta)}{\sin(\beta - \alpha)}.$ Thus, we are done. $\square$

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#19 h Dec 21, 2023, 10:28 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.230909090909087cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.18727272727273, xmax = 15.249090909090896, ymin = -6.133636363636364, ymax = 6.957272727272726; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); draw((-1.97908720787453,1.204050950799106)--(-1.6605867761802946,0.9865243878485139)--(-1.4430602132297026,1.3050248195427494)--(-1.761560644923938,1.5225513824933414)--cycle, linewidth(0.6)); draw((-6.394305392080065,-2.59)--(-6.394305392080065,-2.204305392080065)--(-6.78,-2.204305392080065)--(-6.78,-2.59)--cycle, linewidth(0.6)); /* draw figures */ draw((-6.78,4.95)--(-6.78,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(-4.570314034778992,-2.59), linewidth(0.7) + sexdts); draw((-6.78,-2.59)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-4.570314034778992,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-6.78,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-4.570314034778992,-2.59)--(-1.761560644923938,1.5225513824933414), linewidth(0.7) + sexdts); draw((4.26,-2.59)--(-3.6463196618288167,-1.2370957722321034), linewidth(0.7) + sexdts); /* dots and labels */ dot((-6.78,4.95),dotstyle); label("$B$", (-6.714545454545459,5.139090909090909), NE * labelscalefactor); dot((-6.78,-2.59),dotstyle); label("$A$", (-7.332727272727277,-3.1518181818181823), NE * labelscalefactor); dot((4.26,-2.59),dotstyle); label("$C$", (4.34,-2.4063636363636367), NE * labelscalefactor); dot((-4.570314034778992,-2.59),dotstyle); label("$D$", (-4.423636363636369,-3.17), NE * labelscalefactor); dot((-2.7101665988177412,-1.3972873014084721),dotstyle); label("$E$", (-2.5509090909090975,-1.97), NE * labelscalefactor); dot((-3.6463196618288167,-1.2370957722321034),linewidth(4.pt) + dotstyle); label("$F$", (-3.86,-0.9518181818181822), NE * labelscalefactor); dot((-0.01585517068551292,0.3302851437471739),linewidth(4.pt) + dotstyle); label("$X$", (0.12181818181817432,0.5209090909090904), NE * labelscalefactor); dot((-1.761560644923938,1.5225513824933414),linewidth(4.pt) + dotstyle); label("$G$", (-1.6963636363636432,1.6663636363636358), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Here's a coordinate solution:

Toss the figure in the coordinate plane. Let $G$ be the foot of perpendicular from $D$ to $BC$ . We may assume without loss of generality, that $G = (0,0), \ B = (b,0), \ C = (c,0), \ D = (0,d).$ Now, let $A = (\beta, \alpha)$ . We have $\frac{\alpha}{d} + \frac{\beta}{c} = 1 \ \ \ \ \ \ \ -(1) \ \ \ \ \ \alpha = \frac cd (\beta - b) \ \ \ \ \ \ \ -(2)$ because $A$ lies on $CD$ , and because $A$ also lies on the perpendicular from $B$ to $CD$ . Solving $(1)$ and $(2)$ , we get $\alpha = \frac{cd(c-b)}{(c^2+d^2)}, \ \ \ \beta = \frac{c(d^2+bc)}{(c^2 + d^2)}.$
Now, let $E = (e_x, e_y)$ . The midpoint of $AE$ lies on $BD$ , so $\frac{e_y+\alpha}{2d} + \frac{e_x+\beta}{2b} = 1$ . Also, $AE \perp BD$ , so $e_y - \alpha = \frac{b}{d} \cdot (e_x - \beta)$ . Solving these for $e_x$ , we get (note that $d- \alpha = \frac{d(d^2 + bc)}{(c^2 + d^2)}$ ) $\left(\frac bd + \frac db \right) e_x = \left(\frac bd - \frac db \right) \beta + 2(d- \alpha) \implies \frac{e_x}{\beta} = \frac{b^2 - d^2}{b^2 + d^2} + \frac{2bd(d - \alpha)}{\beta (b^2 + d^2)} = \frac{b^2 - d^2}{b^2 + d^2} + \frac{2bd^2}{c(b^2 + d^2)} = \frac{\gamma}{c(b^2 + d^2)} = \frac{1}{\kappa},$ where $\gamma = c(b^2 - d^2) + 2bd^2$ and $\kappa = \frac{c(b^2 + d^2)}{\gamma}$ .

Now, (using that $\frac{\beta}{\alpha} = \frac{d^2+bc}{d(c-b)}$ ) $\frac{e_y}{\alpha} = 1 + \frac{\beta}{\alpha} \cdot \frac bd \cdot \left( \frac{\gamma}{c(b^2 + d^2)} - 1 \right) = 1+ \frac{b(d^2+bc)}{d^2(c-b)} \cdot \frac{2d^2(b-c)}{c(b^2 + d^2)} = 1 - \frac{2b(d^2 + bc)}{c(b^2 + d^2)} = - \frac{\gamma}{c(b^2 + d^2)} = - \frac{1}{\kappa}.$
Now, let $F = (0,f)$ and let $X = (x,0)$ be the intersection of $DE$ with $BC$ . Then $\frac{e_x}{c} + \frac{e_y}{f} = 1 \implies \frac{\alpha}{f} = \left(1 - \frac{e_x}{c} \right) \frac{\alpha}{e_y} = \left( 1 - \frac{\beta}{c \kappa} \right) (-\kappa) = - \kappa + \frac{\beta}{c}.$ Also, $\frac{e_x}{x} + \frac{e_y}{d} = 1 \implies \frac{\beta}{x} = \left(1 - \frac{e_y}{d} \right) \frac{\beta}{e_x} = \left( 1 + \frac{\alpha}{d \kappa} \right) (\kappa) = \kappa + \frac{\alpha}{d}.$ Thus, it suffices to show that $A$ satisfies the equation of $FX$ , i.e. $\frac{\alpha}{f} + \frac{\beta}{x} = 1 \iff - \kappa + \frac{\beta}{c} + \kappa + \frac{\alpha}{d} = 1$ which is true because of $(1)$ . Thus, we are done. $\square$

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#20 h Dec 21, 2023, 10:50 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.230909090909087cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.18727272727273, xmax = 15.249090909090896, ymin = -6.133636363636364, ymax = 6.957272727272726; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); draw((-1.97908720787453,1.204050950799106)--(-1.6605867761802946,0.9865243878485139)--(-1.4430602132297026,1.3050248195427494)--(-1.761560644923938,1.5225513824933414)--cycle, linewidth(0.6)); draw((-6.394305392080065,-2.59)--(-6.394305392080065,-2.204305392080065)--(-6.78,-2.204305392080065)--(-6.78,-2.59)--cycle, linewidth(0.6)); /* draw figures */ draw((-6.78,4.95)--(-6.78,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(-4.570314034778992,-2.59), linewidth(0.7) + sexdts); draw((-6.78,-2.59)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-4.570314034778992,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-6.78,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-4.570314034778992,-2.59)--(-1.761560644923938,1.5225513824933414), linewidth(0.7) + sexdts); draw((4.26,-2.59)--(-3.6463196618288167,-1.2370957722321034), linewidth(0.7) + sexdts); /* dots and labels */ dot((-6.78,4.95),dotstyle); label("$B$", (-6.714545454545459,5.139090909090909), NE * labelscalefactor); dot((-6.78,-2.59),dotstyle); label("$A$", (-7.332727272727277,-3.1518181818181823), NE * labelscalefactor); dot((4.26,-2.59),dotstyle); label("$C$", (4.34,-2.4063636363636367), NE * labelscalefactor); dot((-4.570314034778992,-2.59),dotstyle); label("$D$", (-4.423636363636369,-3.17), NE * labelscalefactor); dot((-2.7101665988177412,-1.3972873014084721),dotstyle); label("$E$", (-2.5509090909090975,-1.97), NE * labelscalefactor); dot((-3.6463196618288167,-1.2370957722321034),linewidth(4.pt) + dotstyle); label("$F$", (-3.86,-0.9518181818181822), NE * labelscalefactor); dot((-0.01585517068551292,0.3302851437471739),linewidth(4.pt) + dotstyle); label("$X$", (0.12181818181817432,0.5209090909090904), NE * labelscalefactor); dot((-1.761560644923938,1.5225513824933414),linewidth(4.pt) + dotstyle); label("$G$", (-1.6963636363636432,1.6663636363636358), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Here's a projective solution:

Let $G$ be the foot of perpendicular from $D$ to $BC$ and let $X$ be the intersection of $DE$ with $BC$ .

Let $\omega$ be the circle with diameter $BD$ . Then $A \in \omega$ , and therefore the reflection of $A$ across $BD$ is also on $\omega$ , and $(B,D; A,E) = -1$ . Now, $G \in \omega$ therefore $(GB, GD; GA, GE) = -1 = (GC, GD; GA, GE)$ . Hence, $A,F,X$ are collinear by Ceva and Menelaus Theorem in $\displaystyle \triangle DGC$ .

This post has been edited 1 time. Last edited by Rijul saini, Feb 13, 2024, 6:35 AM

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