Solution:

a. Let the quadrilateral be $ABCD$ and let the diagonals $AC$, $BD$ meet at $E$. Then area $ABC = AC \cdot EB \cdot \sin CEB / 2$, and area $ADC = AC \cdot ED \cdot \sin CEB / 2$, so $E$ is the midpoint of $BD$. Similarly, it is the midpoint of $AC$. Hence the triangles $AEB$ and $CED$ are congruent, so $\angle CDE = \angle ABE$, and hence $AB$ is parallel to $CD$. Similarly, $AD$ is parallel to $BC$.

b. Let the hexagon be $ABCDEF$. Let $BE$, $CF$ meet at $J$, let $AD$, $CF$ meet at $K$, and let $AD$, $BE$ meet at $L$. Let $AK = a$, $BJ = b$, $CJ = c$, $DL = d$, $EL = e$, $FK = f$. Also let $KL = x$, $JL = y$ and $JK = z$. Consider the pair of diagonals $AD$, $BE$. They divide the hexagon into 4 parts: the triangles $ALB$ and $DLE$, and the quadrilaterals $AFEL$ and $BCDL$. Since area $ALB$ + area $AFEL$ = area $DLE$ + area $BCDL$, and area $ALB$ + area $BCDL$ = area $DLE$ + area $AFEL$, the two triangles must have the same area (add the two inequalities). But area $ALB = 1/2 \cdot AL \cdot BL \cdot \sin ALB$, and area $DLE = 1/2 \cdot DL \cdot EL \cdot \sin DLE = 1/2 \cdot DL \cdot EL \cdot \sin ALB$, so $AL \cdot BL = DL \cdot EL$ or $de = (a+x)(b+y)$. Similarly, considering the other two pairs of diagonals, we get $bc = (e+y)(f+z)$ and $af = (c+z)(d+x)$. Multiplying the three inequalities gives: $abcdef = (a+f)(b+y)(c+z)(d+x)(e+y)(f+z)$. But $x$, $y$, $z$ are non-negative, so they must be zero and hence the three diagonals pass through a common point.