Answer:
:
Steps: 1 Mark
Proof: 2Marks

In $\Delta 1and\Delta 2$:
$\angle AOD=\angle COD={90}^{\circ }$(Diagonals of square intersect at right angles)
$AD=CD$ (Sides of a square; hypotenuse)
$OD=DO$ (Common)
Hence, $\Delta 1\cong \Delta 2$(By RHS congruence rule) ---------------------1
Similarly, $\Delta 4\cong \Delta 3$(By RHS congruence rule) ---------------------2
In $\Delta 1and\Delta 4$
$\angle AOD=\angle AOB={90}^{\circ }$(Diagonals of square intersect at right angles)
$AD=AB$(Sides of a square; hypotenuse)
$OA=AO$ (Common)
Hence, $\Delta 1\cong \Delta 4$ (By RHS congruence rule) -------------------3
Similarly, $\Delta 2\cong \Delta 3$ (By RHS congruence rule) ----------------4
From 1, 2, 3 and 4 we can say that all the triangles i.e. $\Delta 1$, $\Delta 2$, $\Delta 3$ and $\Delta 4$are congruent to each other.
5, 6, 7 and 8 have relatively small area. Since they have same shape and size, they are also congruent. So, we can say that all the slices of the pizza are congruent to each other.