Answer: Option B. -> OD = OC
:
B
Consider $\Delta BOC$and $\Delta AOD$
1)AD = BC ( Given )
2) $\angle CBO=\angle DAO$= 90°
3) $\angle BOC=\angle AOD$ ......(vertically opposite angles)
$\therefore \Delta BOC\cong \Delta AOD$ ....[AAS Criterion]
$\Rightarrow$ OC = OD ....(congruent parts of congruent triangle)

Answer: Option C. -> Both Statement 1 and statement 2 are true.
:
C
Two line segments are congruent only if they superimpose which is possible only if they have equal length and the converse is also true.
Hence, both the given statementsare true.

Answer: Option A. -> True
:
A
In two triangles,
If a pair of corresponding angles and the included side are equal, then they are congruent [ASA congruence criterion].
If a pair of corresponding angles and a non-included side are equal,then they are congruent [AAS congruence criterion].
Therefore, given statement is true.

Answer: Option A. -> ∠ABD = ∠BAC
:
A
In$\Delta ABD\text{and}\Delta BAC$,
AD = BC(Given)
$\angle BAD$= $\angle CBA$(Given)
AB = BA (Side common to both triangles)
Hence$\Delta ABD\cong \Delta BAC$
(by SAS congruence condition).
Thus, $\angle ABD$ = $\angle BAC$
(congruent parts of congruent triangles).

:
Concept : 1 Mark
Proof : 2 Marks

In $\Delta ACB$ and $\Delta ECD$
$AC=EC$ [Given]
$BC=DC$ [Given]
$\angle ACB=\angle ECD$ [Vertically opposite angles]
$\Rightarrow \Delta ACB\cong \Delta ECD$ (By SAS condition)
$\therefore AB=ED$ [Corresponding parts of congruent triangles]

:
(a) Proof: 2 Marks
(b) Answer: 1 Mark
(a)Consider the two triangles :

In the two triangles,
$\angle ABC$ = $\angle PQR$
$\angle BAC$ = $\angle QPR$
$\angle ACB$ = $\angle PRQ$
But clearly, $\Delta ABC$ is not congruent to $\Delta PQR$.
As the sides of triangle are not equal.
Thus, AAA cannot be a congruence condition.
It actually tells thatthe two triangles are similar, but not congruent.
(b) Since, the three sides of the first triangle is equal to the corresponding three sides of the second triangle, bySSScongruence criterion ΔABCis congruent to ΔDEF.

:
Answer: 1 Mark
Explanation: 2Marks

In $\Delta AEB$ and $\Delta CBD$,
$EB=BD$ [Given]
$AE=CB$ [Given]
$\angle A=\angle C={90}^{\circ }$ [Given]
Hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and one side of another right-angled triangle.
$\therefore \Delta ABE\cong \Delta CDB$ [RHS congruence criterion]

:
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.

:
Each Part: 2 Marks
(a)

In $\Delta BDAand\Delta CDA$
$\angle B=\angle C$ [Given]
$\angle BAD=\angle CAD$ [Given, DA is an angle bisector]
$AD=AD$ [Common side]
$\Rightarrow \Delta BDA\cong \Delta CDA$ [ AAS criteria]
(b) we observethat 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'tenough 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.

:
If two angles have the same measure, they are congruent. Also, if two angles arecongruent, their measures are same.
Hence,the measure of the other angle is also${60}^o$.