In \(\triangle MCL\) and \(\triangle BLC\)
MC = BL
\(\angle MCL=\angle BLC\)
CL = LC
\(\triangle MCL\cong \triangle BLC\)
ML = BC | C.P.C.T
and \(\angle MLC=\angle BCL\) | C.P.C.T
In \(\triangle ABC\) and \(\triangle NML\)
BC = ML
\(\angle BCL=\angle MLN\)
AL = CN
AL + LC = LC + CN
AC = NL
\(\triangle ABC\cong \triangle NML\) | SAS congruence rule

In \(\triangle OAP\) and \(\triangle OBQ\)
AP = BQ
\(\angle OAP=\angle OBQ\)
\(\angle AOB=\angle BOQ\) | Vertically opposite angles
\(\triangle OAP\cong \triangle OBQ\) | AAS Rule
OA = OB | C.P.C.T
OP = OQ | C.P.C.T
O is the mid-point of line segments AB and PQ.

Given: In figure,
ㄥB < ㄥA and ㄥC < ㄥD.
To Prove: AD < BC
Proof: ㄥB < ㄥA
ㄥA > ㄥB
OB> OA ...(1)
ㄥC < ㄥD
ㄥD > ㄥC
OC > OD  ..(2)
| Side opposite to greater angle is longer
From (1) and 2), we get
OB+OC > OA+OD
⇒ BC> AD
⇒ AD < Be.

Given: An equilateral triangle ABC
To prove: \(\angle A+\angle B+\angle C={ 60 }^{ 0 }\)
Proof: ABC is an equilateral triangle
AB = BC = CA ....... (1) | 
AB = BC
\(\angle A=\angle C\) .......... (2) | Angles opposite to equal sides of a triangle are equal
BC = CA
\(\angle A=\angle B\) ......... (3) | Angles opposite to equal sides of a triangle are equal
From (2) and (3), we obtain
\(\angle A=\angle B=\angle C\)  ........ (4)
In \(\triangle ABC\)
\(\angle A+\angle B+\angle C={ 180 }^{ 0 }\) ...... (5) | Sum of all the angles of a triangle is 180°
Let \(\angle A={ x }^{ 0 }\) then, \(\angle B=\angle C={ x }^{ 0 }\)
From (5)
\({ x }^{ 0 }+{ x }^{ 0 }+{ x }^{ 0 }={ 180 }^{ 0 }\)
\(3{ x }^{ 0 }={ 180 }^{ 0 }\)
\({ x }^{ 0 }={ 60 }^{ 0 }\)
\(\angle A=\angle B=\angle C={ 60 }^{ 0 }\)

In \(\triangle\)ABC, AC >AB
\(\therefore\) \(\angle\) ABC > \(\angle\)ACB
(Angles opposite to larger side is greater)
\(\therefore\) \(\angle\)ABC + \(\angle\)1 > \(\angle\)ACB + \(\angle\)1
(Adding \(\angle 1\)on both sides) Y.
\(\therefore\) \(\angle\)ABC + \(\angle\)1 > \(\angle\)ACB + \(\angle\)2
(AD bisects \(\angle\)A, \(\angle\)1 = \(\angle\)2) 
\(\therefore\) \(\angle\)ADC > \(\angle\)ADB.
(Exterior angle property of triangle)

In \(\triangle\)PQS and \(\triangle\)PRS,
PQ = PR  (Given)
PS = PS   (Common)
\(\angle\)PSQ = \(\angle\)PSR = 90°
(PS is altitude)
By R.H.S. rule,
\(\triangle PQS\cong \triangle PRS\)
\(\angle\)QPS = \(\angle\)RPS (By c.p.c.t.) 
Hence, PS bisects \(\angle\)P.

(i) In \(\triangle\) AOD and \(\triangle\)BOC,
OA =OB (Given)
OD = OC (Given)
\(\angle\)AOD = \(\angle\)BOC
(Vertically opposite angles)
So, by SAS criteria,
\(\triangle AOD\cong \triangle BOC\)
(ii) \(\angle\)CBA = \(\angle\)DAB (By c.p.c.t.)
AD and BC are two lines intersected by AB such that \(\angle\)CBA = \(\angle\)DAB and they form a pair of alternate angles.
Hence, AD II BC

In \(\triangle\)PAB and \(\triangle\)PDC, 
PA = PD (Given)
(P is the mid-point of AD)
AB = CD (Side of a square)
\(\angle\)PAB = \(\angle\)PDC = 90°
By R.H.S., \(\triangle PAB\cong \triangle PDC\)
\(\therefore\) PB = PC (By c.p.c.t.)
(Angles opp. to equal sides are equal)
\(\Rightarrow\) \(\angle\)PCB = \(\angle\)PBC. Proved.

BC, AB

AB > BC
ㄥC > ㄥA

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

AB=QR
BC=PR=RP
CA=PQ
∴ A ↔ Q
B ↔ R
C ↔ P
∴ ΔCBA ≅ ΔPRQ

ㄥA+ㄥB+ㄥC=180⁰
⇒ ㄥA+80⁰+ㄥC=180⁰
⇒ ㄥA+ㄥC=100⁰
∵ BC=AB
∴ ㄥA=ㄥC
| Angles opposite to equal sides of a triangle are equal

In ΔAOD and ΔBOC
OA=OB
OD=OC
ㄥAOD=ㄥBOC
∴ ΔAOD≅ΔBOC

Two squares of the same side length ar congruent

↔ denotes correspondence

≌ represents congruence.