#### Answer

$BD = 2$
$AC = 2$

#### Work Step by Step

According to Theorem 6-15, the diagonals of a rectangle are congruent. Therefore, we can set $AC$ and $BD$, the diagonals of $ABCD$, equal to one another to solve for $a$:
$AC = BD$
Substitute with the expressions given for each diagonal:
$2(5a + 1) = 2(a + 1)$
Use the distributive property:
$10a + 2 = 2a + 2$
Subtract $2a$ from each side of the equation to move variables to the left side of the equation:
$8a + 2 = 2$
Subtract $2$ from each side of the equation to move constants to the right side of the equation:
$8a = 0$
Now that we have the value of $a$, we can plug $0$ in for $a$. We do not need to evaluate both expressions because we know that the lengths of the two diagonals are equal.
$BD = 2(0 + 1)$
Evaluate parentheses first:
$BD = 2(1)$
Multiply to solve:
$BD = 2$
If $BD = 2$, then $AC = 2$ because diagonals of rectangles are congruent.