In ΔABC and ΔPQR, we have
AC = PR = 5 cm, BC = PQ = 4 cm
But AB # QR      [∵ 2 cm # 2.5 cm]
So, SSS congruence rule is not applicable.
Hence, ΔABC and ΔPQR are not congruent.

Given, ΔABC = ΔPQR
Also given ㄥB = ㄥQ and ㄥC = ㄥR
[from the given figure)
To apply the condition for congruency, included side of one triangle is equal to the included side of the other triangle.
∴ BC=QR
Hence, we use the ASA congruence criterion.

Area \(\triangle ABC=\frac { 1 }{ 2 } \times AB\times BC=\frac { 1 }{ 2 } \times 4\times 3=6\quad cm^{ 2 }\)
Area \(\triangle EDC=\frac { 1 }{ 2 } \times DE\times CD=\frac { 1 }{ 2 } \times 4\times 3=6\quad cm^{ 2 }\)

 Area of ΔABC = Area of ΔEDC
Also, three sides of MBC are equal to three corresponding sides of ΔEDC, so these two triangles are congruent.
Perimeter of ΔABC = AB + BC + CA = 4 + 3 + 5 = 12 cm
and perimeter of ΔEDC = CD + DE + EC = 3 + 4 + 5 = 12 cm
∴ Perimeter of ΔABC = Perimeter of ΔEDC.

In the given figure, \(\bar { AB } \) and \(\bar { CD } \) bisect each other at O.
So, OC = OD and OA = OB
(i) Three pairs of equal parts in ΔAOC and ΔBOD are OA=OB,OC=OD
and  ㄥAOC = ㄥBOD    [vertically opposite angle]
(ii) In ΔAOC and ΔBOD, we have OA=OB,OC=OD
and ㄥAOC = ㄥBOD [vertically opposite angle]
So, by SAS congruence rule, two triangles are congruent.
The correspondence is A ↔️ B, O ↔️ O and C ↔️ D. In symbolic form, ΔAOC ≅ BOD
Hence, statement (b) i.e. ΔAOC ≅ BOD is true.

(i) Two line segments are congruent, if they have same length.
(ii) Among two congruent angles, one has a measure of 70°, i.ie measure of the other angle is 70°.
(iii) When we write ㄥA = ㄥB, we actually mean mㄥA = mㄥB.

In MQT and b.RSQ, we have
\(\angle\)P= \(\angle\)R, PT= QR, \(\angle\)T= LQ
SO, by SAS congruence rule, two triangles are congruent.
The correspondence is \(P↔️R,\ Q↔️S,\ T↔️Q.\)
In symbolic form, \(\triangle QRS ≅\triangle TPQ\)

In \(Δ\)RAT and \(Δ\)WON, we have
RA = WO [from the given figure]
\(\angle RAT=\angle WON\) [from the given figure]
AT = ON [from the given figure]
So, by SAS congruence rule, two triangles are congruent.
The correspondence is \(A↔️O,\ R↔️W,\ T↔️N.\)
In symbolic form, \(Δ\)RAT \(≅Δ\) WON

All the corresponding congruent parts of  ΔABC and ΔPOR are
\(\angle A ↔️ \angle P, \angle B ↔️ \angle Q, \angle C ↔️ \angle R\) , and \(\overline { AB } \leftrightarrow \overline { PQ } ,\overline { BC } \leftrightarrow \overline { QR } ,\overline { CA } \leftrightarrow \overline { RP } \).

In ΔDAB and ΔCBA,
ㄥDAB = ㄥCBA = 45° + 30° = 75° [given]
AB = AB              [common]
ㄥABD = ㄥBAC = 30° [given]
Therefore, by ASA congruence rule, two triangles are congruent.
The correspondence is A ↔️ B, B ↔️ A and D ↔️ C.
In symbolic form, ΔDAB ≅ ΔCBA

Here, we want to establish that
ΔPQR ≅ ΔFED  [by SAS congruence rule]
Given that, PQ = FE and RP = DF
So, the additional information needed to establish the congruence is ㄥP = ㄥF. 

Given ㄥQPR = 90°, ORT is a line, OP II ST and US II PR, ㄥS=?
Now, we extend US up to U such that it intersects QP at U.
PR || US      [given]
⇒ PR || U'S, U'P is a transversal.

So, ㄥP + ㄥU' = 180° [by the property, sum of two interior angles on the same side of a transversal]
90⁰ + ㄥU = 180°
ㄥU' = 180° - 90°
ㄥU = 90°
Similarly, OP II ST [given]
Then, US is a transversal.
So, ㄥU' + ㄥS =180°  ⇒ 90° + ㄥS =180°
⇒  ㄥS=180⁰ - 90° ⇒  ㄥS = 90°.

According to the question, we have following figure:

Quadrilateral ABCD can be divided into two triangles i.e. ΔABD and ΔBDC.
In two triangles,
AD ≠ CD
AB ≠ BC
BD = BD          [common side]
So, ΔABD ≅ ΔBCD are not true.