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

Given: ABC is an equilateral triangle whose medians are AD, BE and CF.
To prove: AD = BE = CF
Proof: In \(\triangle ADC\) and \(\triangle BEC\)

AC = BC
\(\angle ACD=\angle BCE\)
AD is a median DC = DB = 1/2 BC
BE is a median EA = EC = 1/2 AC
AC = BC
DC = EC
\(\triangle ADC\cong \triangle BEC\) | SAS congruence rule
AD = BE  | C.P.C.T
Similarly, we can prove that
BF = CF ... (2)
CF = AD ....(3)
From (1), (2) and (3)
AD = BE = CF

Given:
AB = FE
BC = ED
\(AB\bot BD\)
\(FE\bot EC\)
To prove: (i) \(\triangle ABD\cong \triangle FEC\)
(ii) \(AD\cong FC\)
Proof: (i) In \(\triangle ABD\) and \(\triangle FEC\)
AB = FE ......... (1)
\(\angle ABD=\angle FEC\) ... (2)   | Each 90⁰
BC = ED
BC + CD = ED + DC
BD = EC ....... (3)
In view of (1), (2) and (3)
\(\triangle ABD\cong \triangle FEC\)  | SAS congruence rule
(ii) AD = FC | C.P.C.T

Given: In a rectangle ABCD, E is a point which bisects BC
To Prove: AE = ED

Proof: In \(\triangle EBA\) and \(\triangle ECD\)
EB = EC
\(\angle EBA=\angle ECD\) | Each 90⁰ (ABCD is a rectangle)
BA = CD | Opposite sides of rectangle ABCD
\(\triangle EBA\cong \triangle ECD\) | SAS congruenece rule
AE = DE | C.P.C.T
AE = ED

(i) AB and CD intersect atO0
\(\therefore\) \(\angle\)AOD = \(\angle\)BOC
(Vertically opp. angles) ...(i)
In \(\triangle\)AOD and \(\triangle\)BOC, we have 
\(\angle\)AOD = \(\angle\)BOC ...(ii)
\(\angle\)DAO = \(\angle\)CBO = 90° (Given)
and AD = BC (Given)
\(\triangle AOD\cong \triangle BOC\)
(By AAS congruence criterion)
\(\Rightarrow\) OA = OB (By c.p.c.t.)
i.e., O is the mid-point of AB
Hence, CD bisects AB.
(ii) Congruency of triangles.
(iii) Equality is the sign of democracy.

(i) I and m are two parallel lines intersected byanother pair of parallel lines p and q.
AD II BC
and AB II CD.
\(\Rightarrow\) ABCD is a parallelogram.
i.e., AB = CD
and BC = AD
Now in \(\triangle\)ABC and \(\triangle\)CDA,we have
AB = CD (Prop. of II^gm)
BC =AD
and AC = AC (Common)
\(\therefore\) \(\triangle ABC\cong \triangle CDA\)
(By SSS criterion of congruence) 
(ii) Congruency of triangles.
(iii) Equality is the sign of democracy.

Proof: QO is bisector of \(\angle\)PQR
\(\angle\)OQR = \(\frac{1}{2}\)\(\angle\)PQR = \(\frac{1}{2}\) =\(\angle\)Q
RO is bisector \(\angle\)ORQ
\(\therefore\) \(\angle\)ORQ =\(\frac{1}{2}\) \(\angle\)PRQ = \(\frac{1}{2}\) \(\angle\)R
In \(\angle\)OQR
\(\angle\)QOR + \(\angle\)OQR + \(\angle\)ORQ = 180°
(Angle sum property)
\(\angle\)QOR + \(\frac{1}{2}\) \(\angle\)Q + \(\frac{1}{2}\) \(\angle\)R = 180°
\(\angle\)QOR = 180°- \(\frac{1}{2}\)(\(\angle\)Q + \(\angle\)R)
But in \(\angle\)PQR
\(\angle\)P + \(\angle\)Q + \(\angle\)R = 180°
\(\angle\)Q + \(\angle\)R = 180°- \(\angle\)P
\(\angle\)QOR = 180°- \(\frac{1}{2}\) (180°- \(\angle\)P)
= 180°-90° + \(\frac{1}{2}\)\(\angle\)P
= 90° + \(\angle\)P Hence Proved.
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In \(\triangle\)ABC, as AB is the greatest side
\(\Rightarrow\) AB > BC \(\Rightarrow\) \(\angle\)C > \(\angle\)A
AB > AC \(\Rightarrow\) \(\angle\)C > \(\angle\)B
On adding (1) and (2), we get
2\(\angle\)C > \(\angle\)A + \(\angle\)B
\(\Rightarrow\) 2\(\angle\)C + \(\angle\)C > \(\angle\)A + \(\angle\)B + \(\angle\)C
\(\Rightarrow\) 3\(\angle\)C > 180°
\(\therefore\) \(\angle\)C > 60°.

Proof: In \(\triangle\)ADE, we have
AD = AE
\(\angle\)ADE = \(\angle\)AED
180°- \(\angle\)ADE = 180°- \(\angle\)AED
\(\Rightarrow\) \(\angle\)ADB = \(\angle\)AEC
Consider \(\triangle\)ABD and \(\triangle\)ACE
AD =AE
\(\angle\)ADB = \(\angle\)AEC
BD = EC
By SAS congruence,
\(\triangle ADB\cong \triangle AEC\)
By c.p.c.t., AB = AC
\(\therefore\) \(\triangle\)ABC is an isosceles triangle.

In \(\triangle\)ABY and \(\triangle\)ACX,
AB = AC   (Given)
AY = AX     (Given)
\(\angle\)A =\(\angle\)A   (Common)
\(\therefore\) By SAS, \(\triangle ABY\cong \triangle ACX\)