All parts: 2 Marks

Given that,
$\mathrm{Δ}PENâ‰\dots \mathrm{Δ}CAR$ and
PN = CR
Corresponding parts of congruent triangle are congruent.
Therefore,  the corresponding sides of congruent triangle are equal.
$⇒PE=CA,\text{Â}\text{Â}\text{Â}EN=AR,\text{Â}\text{Â}\text{Â}\text{Â}PN=CR$.
$⇒$ Also all the corresponding angles of congruent triangles  are equal.
$⇒\mathrm{âˆ}P=\mathrm{âˆ}C,\mathrm{âˆ}E=\mathrm{âˆ}A,\mathrm{âˆ}N=\mathrm{âˆ}R$.

 Reason: 1 Mark each
(a)If three sides of one triangle are equal to the  three sides of the other triangle, then the two  triangles are congruent to each other by SSS congruence criterion.
$⇒$ Both triangles look like the mirror image of each other.
$⇒$ Both the  triangles  superimpose on each other.
So, if the sides of a triangle are congruent to the sides of another triangle, the two triangles will be congruent.
(b) Nothing is given or can be said about any of the corresponding sides in this case, As, AAA is not a rule for congruency, the triangles formed may or may not be congruent, depending on if the corresponding parts are equal or not.

Application of theorem: 1 Mark
Steps: 2 Marks

In $\mathrm{Δ}CAT\text{Â}and\text{Â}\mathrm{Δ}R{A}^{′}T$
$CT=RT$         [Given]
$\mathrm{âˆ}CTA=\mathrm{âˆ}RT{A}^{′}$   [Vertically Opposite Angles]
$AT=A^{′}T$   [Given]
$∴\mathrm{Δ}CATâ‰\dots \mathrm{Δ}R{A}^{′}T$   [By SAS congruence rule]
$⇒\mathrm{âˆ}CAT=\mathrm{âˆ}R{A}^{′}T$   [Corresponding parts of congruent triangles]
But $\mathrm{âˆ}CAT\text{Â}and\text{Â}\mathrm{âˆ}R{A}^{′}T$ are alternate interior angles.
If the pair of alternate interior angles is equal then the lines are parallel.
$⇒CAâˆ\yen A^{′}R$.  

Proof: 1 Mark
Steps: 2 Marks
Consider two triangles, $\mathrm{Δ}ABC$ and $\mathrm{Δ}PQR$ in which,

$\mathrm{âˆ}ABC=\mathrm{âˆ}PQR$
$\mathrm{âˆ}ACB=\mathrm{âˆ}PRQ$
$AB=PQ$
We know that, 
$\mathrm{âˆ}ABC+\mathrm{âˆ}ACB+\mathrm{âˆ}BAC={180}^0$
$\mathrm{âˆ}BAC={180}^0−(\mathrm{âˆ}ABC+\mathrm{âˆ}ACB)....\left(i\right)$
Similarly,
$\mathrm{âˆ}QPR={180}^0−(\mathrm{âˆ}PQR+\mathrm{âˆ}PRQ)......\left(ii\right)$
From (i) and (ii), 
$\mathrm{âˆ}BAC=\mathrm{âˆ}QPR$
Now, In  $\mathrm{Δ}ABC$ and $\mathrm{Δ}PQR$
$\mathrm{âˆ}ABC=\mathrm{âˆ}PQR$  [Given]
$\mathrm{âˆ}BAC=\mathrm{âˆ}QPR$   [ Proved above]
$AB=PQ$  [Given]
$⇒\mathrm{Δ}ABCâ‰\dots \mathrm{Δ}PQR$  [ ASA congruency rule]
$⇒$ These triangles are always congruent.

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.

Properties: 1 Mark
Each proof: 1 Mark

In $\mathrm{Δ}ADB\text{Â}and\text{Â}\mathrm{Δ}ADC$
$AB=AC$    [Given]
$\mathrm{âˆ}ADB=\mathrm{âˆ}ADC={90}^{∘}$   [Given]
$AD=AD$   [common]
Hence, $\mathrm{Δ}ADBâ‰\dots \mathrm{Δ}ADC$ [By RHS congruence rule…….(1)]
From (1), $\mathrm{âˆ}BAD=\mathrm{âˆ}CAD$   [Corresponding parts of congruent triangles]
From (1), $BD=DC$   [Corresponding parts of congruent triangles]

Each Part: 2 Marks
(a)

In $\mathrm{Δ}BDA\text{Â}\text{Â}and\text{Â}\text{Â}\mathrm{Δ}CDA$
$\mathrm{âˆ}B=\mathrm{âˆ}C$    [Given]
$\mathrm{âˆ}BAD=\mathrm{âˆ}CAD$   [Given, DA is an angle bisector ]
$AD=AD$   [Common side]
$⇒\mathrm{Δ}BDAâ‰\dots \mathrm{Δ}CDA$  [  AAS criteria] 
(b) we observe that in the given figures, there are no pairs of congruent sides. Since all of the congruency theorems call for at least one pair of congruent sides, there isn't enough information to prove that the triangles are congruent. Two triangles cannot be proved congruent just by AAA because triangles with same angles can have different sizes.

Properties: 1 Mark
Proof: 1 Mark
Steps: 2 Marks

In a rectangle opposite sides are equal and parallel.
In $\mathrm{Δ}OAD\text{Â}and\text{Â}\mathrm{Δ}OCB$,
$\mathrm{âˆ}ODA=\mathrm{âˆ}OBC$
[Alternate interior angles; $ADâˆ\yen BC$ and BD as transversal]
AD = BC  [Opposite sides of a rectangle are equal]
$\mathrm{âˆ}OAD=\mathrm{âˆ}OCB$  
[Alternate interior angles; $ADâˆ\yen BC$ and AC as transversal]
Hence $\mathrm{Δ}OADâ‰\dots \mathrm{Δ}OCB$   [By ASA congruence rule]
Equating the corresponding parts of congruent triangles, we get:
AO = CO
BO = DO
$⇒$  Diagonals of a rectangle bisect each other.