Let $ABC$ be a right triangle with $m(\widehat{A})=90^\circ$. Let $APQR$ be a square with area $9$ such that $P\in [AC]$, $Q\in [BC]$, $R\in [AB]$. Let $KLMN$ be a square with area $8$ such that $N,K\in [BC]$, $M\in [AB]$, and $L\in [AC]$. What is $|AB|+|AC|$?

$ 
\textbf{(A)}\ 8
\qquad\textbf{(B)}\ 10
\qquad\textbf{(C)}\ 12
\qquad\textbf{(D)}\ 14
\qquad\textbf{(E)}\ 16
$
1. Given that \( \triangle ABC \) is a right triangle with \( \angle A = 90^\circ \), and \( APQR \) is a square with area 9 such that \( P \in [AC] \), \( Q \in [BC] \), \( R \in [AB] \).
2. Since the area of square \( APQR \) is 9, the side length of the square is \( \sqrt{9} = 3 \).
3. Let \( AQ \) be the angle bisector of \( \triangle ABC \). The length of \( AQ \) is \( 3\sqrt{2} \) because the diagonal of a square with side length 3 is \( 3\sqrt{2} \).
4. Let \( AH \) be the altitude from \( A \) to \( BC \). Since \( \text{Angle Bisector} \geq \text{Median} \geq \text{Altitude} \), we have \( AQ = 3\sqrt{2} \geq \frac{BC}{2} > AH \).
5. Therefore, \( BC \leq 6\sqrt{2} \) and \( AH \leq 3\sqrt{2} \).
6. Let \( KLMN \) be a square with area 8 such that \( N, K \in [BC] \), \( M \in [AB] \), and \( L \in [AC] \). The side length of this square is \( \sqrt{8} = 2\sqrt{2} \).
7. Let \( ML \) cut \( AH \) at \( S \). Since \( SH = 2\sqrt{2} \), we have \( AS \leq \sqrt{2} \).
8. Using the similarity of triangles, \( \frac{AS}{AH} = \frac{ML}{BC} \Rightarrow \frac{AS}{SH} = \frac{ML}{BC - ML} \).
9. Therefore, \( ML \cdot SH = AS \cdot (BC - ML) \). Given \( AS \leq \sqrt{2} \) and \( BC - ML \leq 4\sqrt{2} \), the left-hand side (LHS) is \( 2\sqrt{2} \cdot 2\sqrt{2} = 8 \), and the right-hand side (RHS) is \( \leq \sqrt{2} \cdot 4\sqrt{2} = 8 \).
10. Since both sides are equal, equality occurs when \( Q = H \). Thus, \( \triangle ABC \) is isosceles, meaning \( AB = AC = 6 \).
11. Therefore, \( AB + AC = 6 + 6 = 12 \).

The final answer is \( \boxed{12} \).