In triangle $ ABC$, $ AB\equal{}AC$, the inscribed circle $ I$ touches $ BC, CA, AB$ at points $ D,E$ and $ F$ respectively. $ P$ is a point on arc $ EF$ opposite $ D$. Line $ BP$ intersects circle $ I$ at another point $ Q$, lines $ EP$, $ EQ$ meet line $ BC$ at $ M, N$ respectively. Prove that
(1) $ P, F, B, M$ concyclic
(2)$ \frac{EM}{EN} \equal{} \frac{BD}{BP}$

(P.S. Can anyone help me with using GeoGebra, the incircle function of the plugin doesn't work with my computer.)
### Part 1: Prove that \( P, F, B, M \) are concyclic

1. **Identify the harmonic quadrilateral:**
   Since \( QFPD \) is a harmonic quadrilateral, it implies that the cross-ratio \((Q, F; P, D)\) is -1. This property will be useful in our proof.

2. **Use the harmonic property:**
   The harmonic property implies that \( MD = DN \). Therefore, \( P, F, N \) are collinear. This collinearity will help us in angle chasing.

3. **Angle chasing:**
   Since \( P, F, N \) are collinear, we can use the angles to show that \( P, F, B, M \) are concyclic. We need to show that \(\angle PFB = \angle PMB\).

4. **Symmetry and angle chasing:**
   By symmetry and simple angle chasing, we can show that \(\angle FBP = \angle PMF\). This implies that the points \( P, F, B, M \) lie on a circle.

Thus, \( P, F, B, M \) are concyclic.

### Part 2: Prove that \( \frac{EM}{EN} = \frac{BD}{BP} \)

1. **Harmonic quadrilateral property:**
   Recalling the property of the harmonic quadrilateral \( QFBD \), we obtain that \( BD = BN \).

2. **Similar triangles:**
   Simple angle chasing yields that \( \triangle BPN \sim \triangle EMN \). This similarity will help us in finding the ratio of the sides.

3. **Ratio of sides:**
   From the similarity of triangles \( \triangle BPN \) and \( \triangle EMN \), we have:
   \[
   \frac{EM}{EN} = \frac{BP}{BN}
   \]
   Since \( BD = BN \), we can substitute \( BN \) with \( BD \):
   \[
   \frac{EM}{EN} = \frac{BP}{BD}
   \]
   Taking the reciprocal, we get:
   \[
   \frac{EM}{EN} = \frac{BD}{BP}
   \]

Thus, we have proved that \( \frac{EM}{EN} = \frac{BD}{BP} \).

The final answer is \( \boxed{\frac{EM}{EN} = \frac{BD}{BP}} \)