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### Easy Projective Geo

Posted: Tue Apr 17, 2018 10:17 am
Points $A, B, C, D, E$ lie on a circle and a point $P$ lie outside the circle. The given points such that (1) lines $PB$ and $PD$ are tangent to the circle, (2) $P, A, C$ are collinear and (3) $DE$ is parallel to $AC$.
Prove that: $BE$ bisects $AC$

### Re: Easy Projective Geo

Posted: Wed Apr 18, 2018 6:31 pm
We get $ACDE$ as an isosceles trapezoid
Angle chasing gives us that
$\angle{ABE}=\angle{DBC}$
Let $Q$ be the intersection point of $AC$ and $BD$.
Again, $(A,C;Q,P)$ is a harmonic bundle
$\frac{AQ}{CQ}=\frac{AP}{CP}$
Now it is enough to show that $BD$ is a symmedian
Which means to show, $\frac{AB^2}{BC^2}=\frac{AQ}{CQ}=\frac{AP}{CP}$
$\frac{AB^2}{AP}=\frac{BC^2}{CP}$
$\frac{AB\sin{\angle{BPC}}}{\sin{\angle{ABP}}}=\frac{BC\sin{\angle{BPC}}}{\sin{\angle{CBP}}}$
$\frac{AB}{\sin{\angle{ACB}}}=\frac{BC}{\sin{\angle{BAC}}}$
Which is obvious by sine law...

### Re: Easy Projective Geo

Posted: Thu Apr 19, 2018 9:35 am
4.PNG (37.17 KiB) Viewed 4564 times
We will use this lemma:

Lemma: A point $P$ is outside or on a circle $\omega$. Let $PC$ and $PD$ be tangents to $\omega$, and $\iota$ be a line through $P$ intersecting $\omega$ at $A,B$. Let $AB$ intersect $CD$ at $Q$. Then $ABCD$ is a harmonic quadrilateral and $(P,Q;A,B)$ is harmonic.

We have: $P$ is outside a circle $\omega$. $PB$ and $PC$ are tangents to $\omega$. $P,A,C$ are collinear. $BE$ and $AC$ intersect at point $M$, $BD$ and $AC$ intersect at $N$. So, $P,A,N,M,C$ are collinear. According to the lemma:
(1) $ABCD$ is a harmonic quadrilateral, (2) $(P,N;A,C)$ is harmonic. So,
$\frac{PA}{AN}:\frac{PC}{NC}=1$
$\Rightarrow \frac{PA}{AN}=\frac{PC}{NC} \Rightarrow \frac{PA}{PC}=\frac{AN}{NC}$

Let $PA=a,AN=b,NM=c,MC=d$. So we get:
$\frac {a}{a+b+c+d}=\frac{b}{c+d} \Rightarrow ac+ad=ab+b^2+bc+bd$ ...(1)
$AC$ is parallel to $DE$. So, $\angle ACD=\angle CDE$. But $\angle PDA=\angle ACD, \angle CDE=\angle CBE$. So we get: $\angle PDA=\angle CBE$
On arc $AB$, $\angle ADB=\angle ACB$.
So, $\angle PDB=\angle PDA+\angle ADB=\angle CBE+\angle ACB= \angle BMP$. So, $P,D,M,B$ are cyclic.

Power of $N$ with respect to $\bigcirc PDMB$, $PN \times PM=BN \times ND$
Power of $N$ with respect to $\bigcirc ABCD$, $BN \times ND=AN \times NC$
So, $PN \times PM=AN \times NC$
$\Rightarrow (a+b)c=(c+d)b$
$\Rightarrow ac+bc=bc+bd$
$\Rightarrow a=\frac {bd}{c}$

Applying the value of $a$ to the equation (1),
$ac+bd=ab+b^2+bc+bd$
$\Rightarrow bcd+bd^2=b^d+b^c+bc^2+bcd$
$\Rightarrow bd^2=b^2d+b^2c+bc^2$
$\Rightarrow bd^2-bc^2=b^2d+b^2c$
$\Rightarrow b(d+c)(d-c)=b^2(d+c)$
$\Rightarrow b=d-c$
$\Rightarrow d=b+c$
So, $MC=AN+NM$
$Q.E.D$

### Re: Easy Projective Geo

Posted: Fri Apr 20, 2018 9:59 pm
Harmonic quad approach is quite intuitive. Anyone tried bash?
Btw, apart from projective I solved it by cartesian coordinates. I don't have enough patience to type that lengthy and annoying solution here.