Steps: 1 Mark
Proof: 2 Marks

In $\mathrm{Δ}1\text{Â}and\text{Â}\mathrm{Δ}2$:
$\mathrm{âˆ}AOD=\mathrm{âˆ}COD={90}^{∘}$ (Diagonals of square intersect at right angles)
$AD=CD$ (Sides of a square; hypotenuse)
$OD=DO$ (Common)
Hence,  $\mathrm{Δ}1â‰\dots \mathrm{Δ}2$ (By RHS congruence rule) ---------------------1
Similarly, $\mathrm{Δ}4â‰\dots \mathrm{Δ}3$ (By RHS congruence rule) ---------------------2
In $\mathrm{Δ}1\text{Â}and\text{Â}\mathrm{Δ}4$
$\mathrm{âˆ}AOD=\mathrm{âˆ}AOB={90}^{∘}$ (Diagonals of square intersect at right angles)
$AD=AB$ (Sides of a square; hypotenuse)
$OA=AO$ (Common)
Hence, $\mathrm{Δ}1â‰\dots \mathrm{Δ}4$ (By RHS congruence rule) -------------------3
Similarly, $\mathrm{Δ}2â‰\dots \mathrm{Δ}3$ (By RHS congruence rule) ----------------4
From 1, 2, 3 and 4 we can say that all the triangles i.e. $\mathrm{Δ}1$, $\mathrm{Δ}2$, $\mathrm{Δ}3$ and $\mathrm{Δ}4$ 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.