201. Applications of Desarques' and Pappus' theorems.

by Virgil Nicula, Dec 30, 2010, 7:52 AM

Desarques' theorem. Let $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ for which denote $\left\|\begin{array}{c}
M\in B_1C_1\cap B_2C_2\\\
N\in C_1A_1\cap C_2A_2\\\
P\in A_1B_1\cap A_2B_2\end{array}\right\|$ . Then $A_1A_2\cap B_1B_2\cap C_1C_2\ne\emptyset\ \iff\ P\in MN$ .

Proof. Suppose that $X\in A_1A_2\cap B_1B_2\cap C_1C_2$ . Apply the Menelaus' theorem to the following transversals in the mentioned triangles :

$\left\|\begin{array}{cc}
\overline {MB_2C_2}/\triangle XB_1C_1\ : & \frac {MB_1}{MC_1}\cdot\frac {C_2C_1}{C_2X}\cdot\frac {B_2X}{B_2B_1}=1\\\\
\overline {NC_2A_2}/\triangle XC_1A_1\ : & \frac {NC_1}{NA_1}\cdot\frac {A_2A_1}{A_2X}\cdot\frac {C_2X}{C_2C_1}=1\\\\
\overline {PA_2B_2}/\triangle XA_1B_1\ : & \frac {PA_1}{PB_1}\cdot\frac {B_2B_1}{B_2X}\cdot\frac {A_2X}{A_2A_1}=1\end{array}\right\|\ \ \bigodot$ $\implies$ $\frac {MB_1}{MC_1}\cdot\frac {NC_1}{NA_1}\cdot\frac {PA_1}{PB_1}=1$ $\iff\ P\in MN$ . Suppose that $P\in MN$ and denote

$X\in A_1A_2\cap B_1B_2$ . Observe that $\triangle MB_1B_2$ and $\triangle NA_1A_2$ have the property $P\in MN\cap B_1A_1\cap B_2A_2$ . From the direct implication obtain that

$C_1\in MB_1\cap NA_1$ , $C_2\in MB_2\cap NA_2$ and $X\in B_1B_2\cap A_1A_2$ are colinearly, i.e. $X\in C_1C_2$ what means that $A_1A_2\cap B_1B_2\cap C_1C_2\ne\emptyset$ .

Remark. The triangles $A_1B_1C_1$ and $A_2B_2C_2$ in this case are named homologically, the point $X$ is the homology center and $\overline{MNP}$

is named homology line. In particular case $A_1A_2\parallel B_1B_2\parallel C_1C_2$ , i.e. $X\rightarrow\infty$ , also the points $M$ , $N$ , $P$ are colinearly.



Pappus' theorem. Let $d_1$ , $d_2$ be two lines (concurrent or parallel) and $\left\|\begin{array}{c}
\left\{A_1,B_1,C_1\right\}\subset d_1\\\
\left\{A_2,B_2,C_2\right\}\subset d_2\end{array}\right\|$ . Denote $\left\|\begin{array}{c}
X\in B_1C_2\cap B_2C_1\\\
Y\in C_1A_2\cap C_2A_1\\\
Z\in A_1B_2\cap A_2B_1\end{array}\right\|$ . Then $Z\in XY$ .

Proof. Suppose that $d_1\cap d_2\ne\emptyset$ and denote $O\in d_1\cap d_2$ , $A\in XZ\cap d_2$ , $B\in XY\cap d_2$ . Apply the Menelaus' theorem to the following transversals in the

mentioned triangles : $\left\|\begin{array}{ccc}
\overline{AZX}/\triangle A_2B_1C_2\ : & \frac {AC_2}{AA_2}\cdot \frac {ZA_2}{ZB_1}\cdot \frac {XB_1}{XC_2}=1 & (1)\\\\
\overline{B_2ZA_1}/\triangle OA_2B_1\ : & \frac {B_2A_2}{B_2O}\cdot\frac {A_1O}{A_1B_1}\cdot\frac {ZB_1}{ZA_2}=1 & (2)\\\\
\overline{B_2XC_1}/\triangle OB_1C_2\ : & \frac {C_1B_1}{C_1O}\cdot\frac {B_2O}{B_2C_2}\cdot\frac {XC_2}{XB1}=1 & (3)\end{array}\right\|\ \bigodot\implies$ $\frac {AA_2}{AC_2}=\frac {B_2A_2}{B_2C_2}\cdot \frac {A_1O}{C_1O}\cdot\frac {C_1B_1}{A_1B_1}$ . Obtain analogously

$\left(B_1\rightarrow C_1\ ,\ B_2\rightarrow C_2\right)$ $\frac {BA_2}{BB_2}=\frac {C_2A_2}{C_2B_2}\cdot\frac {A_1O}{B_1O}\cdot\frac {B_1C_1}{A_1C_1}$ . From the last two relations obtain that $\frac {AA_2}{AC_2}\cdot\frac {BB_2}{BA_2}=\frac {B_2A_2}{B_2C_2}\cdot\frac {A_1C_1}{C_2A_2}$ ,

where $\left\{A,B,A_2,B_2,C_2\right\}\subset d_2$ $\implies$ $\frac {BA_2}{BB_2}=\frac {AA_2}{AB_2}$ $\implies A\equiv B\implies Z\in XY$ .

Remark. Can prove analogously the case $d_1\parallel d_2$ applying the Thales' theorem for the relations $(2)$ and $(3)$ . The Pappus' theorem

is a limit case of the Desarques' theorem and can adapt the proof of the Desarques' theorem to prove the Pappus' theorem.



PP1. Let $ABC$ be a triangle and two points $P$, $Q$ which belong to the inside of $\triangle ABC$.

I denote $\left\|\begin{array}{cc}
X_a\in BQ\cap CP & Y_a\in BP\cap CQ\\\
X_b\in CP\cap AQ & Y_b\in CQ\cap AP\\\
X_c\in AP\cap BQ & Y_c\in AQ\cap BP\end{array}\right\|$ . Prove that $X_aY_a\cap X_bY_b\cap X_cY_c\ne\emptyset$ .


Proof. I denote the intersections $U$, $V$, $W_x$, $W_y$ of the line $BC$ with the lines $AP$, $AQ$, $X_bX_c$, $Y_bY_c$ respectively. I"ll apply Menelaus' theorem

for $\triangle AUV$ and the transversals $\left\|\begin{array}{ccc}
\overline {BX_cQ}\implies \frac{BU}{BV}\cdot \frac{QV}{QA}\cdot \frac{X_cA}{X_cU}=1 & ; & \overline {BY_cP}\implies \frac{BV}{BU}\cdot \frac{Y_cA}{Y_cV}\cdot \frac{PU}{PA}=1\\\\
\overline {CX_bP}\implies \frac{CU}{CV}\cdot \frac{X_bV}{X_bA}\cdot \frac{PA}{PU}=1 & ; & \overline {CY_bQ}\implies \frac{CV}{CU}\cdot \frac{QA}{QV}\cdot \frac{Y_bU}{Y_bA}=1\\\\
\overline {X_bX_cW_x}\implies \frac{W_xV}{W_xU}\cdot \frac{X_cU}{X_cA}\cdot \frac{X_bA}{X_bV}=1 & ; &  \overline {Y_bY_cW_y}\implies \frac{W_yU}{W_yV}\cdot \frac{Y_bA}{Y_bU}\cdot \frac{Y_cV}{Y_cA}=1\end{array}\right\|$ .

From the product of the above relations we obtain the relation $\frac{W_xU}{W_xV}=\frac{W_yU}{W_yV}$ , i.e. $W_x\equiv W_y\equiv W$ $\implies$ $X_bX_c\cap Y_bY_c\cap BC\ne\emptyset$ (in the point $W$).

For $\triangle X_aX_bX_c$ and $\triangle Y_aY_bY_c$ we observe that $\left\|\begin{array}{c}
C\in X_aX_b\cap Y_aY_b\\\
B\in X_aX_c\cap Y_aY_c\\\
W\in X_bX_c\cap Y_bY_c\end{array}\right\|$ and $W\in BC$ . From Desarques' theorem obtain that $X_aY_a\cap X_bY_b\cap X_cY_c\ne\emptyset$ .

Some nice and interesting remarks.

$\blacktriangleright\ A\in X_aY_a\ \vee \ B\in X_bY_b\ \vee \ C\in X_cY_c\Longleftrightarrow W\in PQ$ . Indeed, can apply the Desarques' theorem

to $\triangle PX_bY_c$ and $\triangle QX_cY_b$ , where $\left\|\begin{array}{c}
X_a\in PX_b\cap QX_c\\\
Y_a\in PY_c\cap QY_b\\\
A\in X_bY_c\cap X_cY_b\end{array}\right\|$ . Thus, $A\in X_aY_a$ $\Longleftrightarrow$ $PQ\cap X_bX_c\cap Y_bY_c\ne \emptyset$ .

$\blacktriangleright$ Here is a solution using the Pappus' theorem. Indeed, $\{Y_b,P,X_c\}\subset AP$ , $\{Y_c,Q,X_b\}\subset AQ\Longrightarrow$ the points $Y_a\in Y_bQ\cap PY_c$ ,

$X_a\in PX_b\cap X_cQ$ and the intersection of the lines $Y_bX_b$, $X_cY_c$ are collinearly, i.e. the lines $X_aY_a\cap X_bY_b\cap X_cY_c\ne\emptyset$ .



PP2. Let $ABC$ be a triangle with the incircle $i=C(I,r)$ which touches its sides in the points $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ . I denote

second intersections $M$ , $N$ , $P$ of $AI$ , $BI$ , $CI$ respectively with the circumcircle $e=C(O,R)$ and the second intersections $A'$ , $B'$ , $C'$ of $AO$ , $BO$ , $CO$

respectively with circle $e$ . Prove that $MD\cap NE\cap PF\ne\emptyset$ and the points $U\in MD\cap A'I$ , $V\in NE\cap B'I$ , $V\in PF\cap C'I$ belong to the circle $e$.


Proof. $\left\|\begin{array}{ccc}
PB=PI\ ,\ MB=MI & \Longrightarrow & PM\perp BI\\\
BF=BD & \Longrightarrow & FD\perp BI\end{array}\right\|$ $\implies$ $PM\parallel FD$ . Prove analogously $\begin{array}{c}
 MN\parallel DE\\\
 NP\parallel EF\end{array}\Longrightarrow$ (Desarques' theorem)

$MD\cap NE\cap PF\ne \emptyset$ . Denote $S\in MD\cap A'I$ . Observe that $\widehat {DIM}\equiv \widehat {IAA'},\ \frac{ID}{IM}=\frac{AI}{AA'}$ because $2Rr=ID\cdot AA'$

and $IA\cdot IM=p(I)$ - the power of the point $I$ w.r.t the circumcircle $e$ $\Longrightarrow$ $\triangle DIM\sim \triangle IAA'\Longrightarrow \widehat {IMD}\equiv \widehat {AA'I}\Longrightarrow S\in e$ .



PP3. For the parallelogram $ABCD$ denote $\left\{\begin{array}{c}EG\parallel AD\mathrm{,\ where\ }E\in AB\ ,\ G\in CD\\\\ FH\parallel AB\mathrm{,\ where\ }F\in BC\ ,\ H\in AD\end{array}\right\|$ and $\left\{\begin{array}{c}J\in EG\cap FH\\\ K\in CH\cap AG\end{array}\right\|$ . Prove that $B\in JK$ .

Proof 1. Apply the direct Menelaus' theorem to the transversal $\overline{AKG}$ in $\triangle CDH\ :\ \frac{AH}{AD}\cdot\frac{GD}{GC}\cdot\frac{KC}{KH}=1$ .

From $\left\{\begin{array}{c}AH=BF\ ,\ AD=BC\\\ GD=JH\ ,\ GC=JF\end{array}\right\|$ obtain that $\frac{BF}{BC}\cdot \frac{KC}{KH}\cdot\frac{JH}{JF}=1$, where $B\in FC$, $K\in CH$, $J\in HF$.

From the Menelaus' theorem applied in $\triangle CFH$ obtain $B\in JK$. This problem is a particular case of the Brianchon's theorem.

Proof 2 (only use Thales' theorem). Let $M\in AG\cap CD$ and $N\in BJ\cap CD$ . Therefore,

$\frac{NG}{CG}=\frac{NG}{BE}=$ $\frac{JG}{JE}=$ $\frac{JG}{HA}=\frac{JM}{HM}$ and $HM\parallel CG\implies K\in JN\implies B\in JK$ .

Remark. This problem is a particular case of the Brianchon's theorem.


The Brianchon's theorem. Let $ABCD$ be a convex quadrilateral for which denote $\left\{\begin{array}{c}M\in AD\cap BC\\\ N\in AB\cap CD\end{array}\right\|$ . Consider

four points $\left\{\begin{array}{c}E\in AB\ ,\ F\in BC\\\ G\in CD\ ,\ H\in DA\end{array}\right\|$ for which so that $\left\{\begin{array}{c}M\in EG\\\ N\in FH\end{array}\right\|$ . Define $\left\{\begin{array}{c}J\in EG\cap FH\\\ K\in AG\cap CH\end{array}\right\|$. Then $B\in JK$ .


Proof. Apply the Pappus' theorem to the lines $\overline{MAH}$ and $\overline{NCG}\ :\ \left\{\begin{array}{c}B\in MC\cap AN\\\ J\in MG\cap HN\\\ K\in AG\cap HC\end{array}\right\|$ $\implies$ $B\in JK$ .


PP4. In the parallelogram $ABCD$ for $E\in (AB)$ and $F\in (CD)$ define $\begin{array}{c}
K\in DE\cap AF\\\
M\in KL\cap AD\end{array}$ and $\begin{array}{c}
L\in CE\cap BF\\\
N\in KL\cap BC\end{array}$ . Prove that $AM = CN$ .

Proof. By Pappus' theorem obtain that $O\in AC\cap BD\cap MN\ne\emptyset$ . In conclusion, the result follows by symmetry.


Pascal's theorem. Let $A_1A_2A_3A_4A_5A_6$ be a cyclical hexagon for which denote $\left\{\begin{array}{c}
M\in A_1A_2\cap A_4A_5\\\
N\in A_2A_3\cap A_5A_6\\\
P\in A_3A_4\cap A_6A_1\end{array}\right\|$ . Prove that $P\in MN$ .

Proof. Denote the intersections $\left\{\begin{array}{c}
A\in A_2A_3\cap A_4A_5\\\
B\in A_4A_5\cap A_6A_1\\\
C\in A_6A_1\cap A_2A_3\end{array}\right\|$ . Using the power of the points $(A,B,C)$ w.r.t. the circumscribed circle obtain that $\left\{\begin{array}{ccc}
AA_2\cdot AA_3 & = & AA_4\cdot AA_5\\\
BA_4\cdot BA_5 & = & BA_6\cdot BA_1\\\
CA_6\cdot CA_1 & = & CA_2\cdot CA_3\end{array}\right\|\ (*)$ .

Apply the Manelaus' theorem to thr triangle $ABC$ and the transversals $:\ \left\{\begin{array}{ccc}
\overline{MA_1A_2}\ : & \implies & \frac {MB}{MA}\cdot\frac {A_2A}{A_2C}\cdot\frac {A_1C}{A_1B}=1\\\\
\overline{NA_5A_6}\ : & \implies & \frac {NA}{NC}\cdot\frac {A_6C}{A_6B}\cdot\frac {A_5B}{A_5A}=1\\\\
\overline{PA_3A_4}\ : & \implies & \frac {PC}{PB}\cdot\frac {A_4B}{A_4A}\cdot\frac {A_3A}{A_3C}=1\end{array}\right\|\ \bigodot\ \stackrel{(*)}{\implies}\ \frac {MB}{MA}$ $\cdot\frac {NA}{NC}\cdot\frac {PC}{PB}=1\implies$ $P\in MN$ .
This post has been edited 83 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:21 PM

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