**(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 $AC$**

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 $AC$**

Prove that:

- Ananya Promi
**Posts:**36**Joined:**Sun Jan 10, 2016 4:07 pm**Location:**Naogaon, Bangladesh

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...

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...

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:

$\frac{PA}{AN}:\frac{PC}{NC}=1$

$\Rightarrow \frac{PA}{AN}=\frac{PC}{NC} \Rightarrow \frac{PA}{PC}=\frac{AN}{NC}$

Let $PA=a,AN=b,NM=c,MC=d$. So we get:

$\frac {a}{a+b+c+d}=\frac{b}{c+d} \Rightarrow ac+ad=ab+b^2+bc+bd$ ...

$AC$ 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

$ac+bd=ab+b^2+bc+bd$

$\Rightarrow bcd+bd^2=b^d+b^c+bc^2+bcd$

$\Rightarrow bd^2=b^2d+b^2c+bc^2$

$\Rightarrow bd^2-bc^2=b^2d+b^2c$

$\Rightarrow b(d+c)(d-c)=b^2(d+c)$

$\Rightarrow b=d-c$

$\Rightarrow d=b+c$

So, $MC=AN+NM$

$Q.E.D$

- Atonu Roy Chowdhury
**Posts:**64**Joined:**Fri Aug 05, 2016 7:57 pm**Location:**Chittagong, Bangladesh

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.

Btw, apart from projective I solved it by cartesian coordinates. I don't have enough patience to type that lengthy and annoying solution here.

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