(1) $D$ is an arbitary point in $\triangle{ABC}$. Prove that:
\[ \frac{BC}{\min{AD,BD,CD}} \geq \{ \begin{array}{c} \displaystyle 2\sin{A}, \ \angle{A}< 90^o \\ \\ 2, \ \angle{A} \geq 90^o \end{array}  \]
(2)$E$ is an arbitary point in convex quadrilateral $ABCD$. Denote $k$ the ratio of the largest and least distances of any two points among $A$, $B$, $C$, $D$, $E$. Prove that $k \geq 2\sin{70^o}$. Can equality be achieved?
### Part (1)

Given: \( D \) is an arbitrary point in \( \triangle ABC \).

We need to prove:
\[ \frac{BC}{\min{AD, BD, CD}} \geq \begin{cases} 
2\sin{A}, & \text{if } \angle{A} < 90^\circ \\
2, & \text{if } \angle{A} \geq 90^\circ 
\end{cases} \]

1. **Case 1: \( \angle A < 90^\circ \)**

   - By the Law of Sines in \( \triangle ABD \):
     \[
     \frac{BD}{\sin A} = \frac{AD}{\sin \angle ABD}
     \]
     Since \( \sin \angle ABD \leq 1 \), we have:
     \[
     BD \leq AD \cdot \sin A
     \]
     Similarly, in \( \triangle ACD \):
     \[
     \frac{CD}{\sin A} = \frac{AD}{\sin \angle ACD}
     \]
     Since \( \sin \angle ACD \leq 1 \), we have:
     \[
     CD \leq AD \cdot \sin A
     \]
     Therefore, the minimum distance from \( D \) to the sides \( BC, CA, AB \) is at most \( AD \cdot \sin A \). Hence:
     \[
     \min(AD, BD, CD) \leq AD \cdot \sin A
     \]
     Thus:
     \[
     \frac{BC}{\min(AD, BD, CD)} \geq \frac{BC}{AD \cdot \sin A} \geq 2 \sin A
     \]

2. **Case 2: \( \angle A \geq 90^\circ \)**

   - In this case, \( \sin A \geq 1 \). Therefore:
     \[
     \frac{BC}{\min(AD, BD, CD)} \geq \frac{BC}{AD} \geq 2
     \]

Thus, the inequality holds in both cases.

### Part (2)

Given: \( E \) is an arbitrary point in convex quadrilateral \( ABCD \). Denote \( k \) as the ratio of the largest and least distances of any two points among \( A, B, C, D, E \).

We need to prove:
\[ k \geq 2\sin{70^\circ} \]

1. **Assume \( E \) lies on \( AC \) or \( BD \)**

   - If \( E \) lies on \( AC \) or \( BD \), then the maximum distance is at least the length of the diagonal, and the minimum distance is at most half the length of the diagonal. Hence:
     \[
     k \geq 2 > 2 \sin 70^\circ
     \]

2. **Assume \( E \) is in the strict interior of \( \triangle BCD \) and \( \triangle BCA \)**

   - If any triangle consisting of three of \( A, B, C, D, E \) has an obtuse angle greater than \( 140^\circ \), we are done by the Law of Sines within that triangle.
   - By part (1), if any of \( \angle BAC, \angle ABC, \angle ACB, \angle BDC, \angle DCB, \angle DBC \) are greater than or equal to \( 70^\circ \), we are done.
   - Suppose all angles are in the range \( (40^\circ, 70^\circ] \). Let \( \theta_1 = \angle ABC \) and \( \theta_2 = \angle DCB \).
   - If \( \angle BAD \leq 70^\circ \), then \( \angle ADC \geq 150^\circ \) and we are done. Similarly, if \( \angle ADC \leq 70^\circ \), we are done.
   - Assume \( \angle BAD, \angle CDA > 70^\circ \). Either \( \angle BAD \leq 180^\circ - \theta_1 \) or \( \angle ADC \leq 180^\circ - \theta_2 \). Consider \( \angle BAD \leq 180^\circ - \theta_1 \).
   - We have \( \angle BAD \in [70^\circ, 180^\circ - \theta_1] \). Since \( \theta_1 \leq 70^\circ \), \( \sin \angle BAD \geq \sin \theta_1 \).
   - Therefore:
     \[
     \frac{BD}{AD} \geq \frac{\sin \theta_1}{\sin (\theta_1 - 40^\circ)}
     \]
     where \( \angle ABD = \angle ABC - \angle DBC \leq \theta_1 - 40^\circ \).
   - For \( \theta_1 \in (40^\circ, 70^\circ] \):
     \[
     \frac{\sin \theta_1}{\sin (\theta_1 - 40^\circ)} \geq \frac{\sin 70^\circ}{\sin 30^\circ} = 2 \sin 70^\circ
     \]
     Hence, \( k \geq 2 \sin 70^\circ \).

Equality can be achieved if the points are positioned such that the distances align with the bounds given by the Law of Sines.

The final answer is \( \boxed{ k \geq 2\sin{70^\circ} } \)