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**I just made it harder**!

**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, $MC=AN+NM$

$Q.E.D$