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 and CD are perpendicular to the line segment AD. AD and BC intersect at P such that PA = PD.
To Prove: (i) AB = CD
(ii) P is the mid-point of BC.
Proof: (i) In \(\triangle ABD\) and \(\triangle PDC\)
PA = PD
\(\angle APB=\angle DPC\) | Vertically opposite angles
\(\angle PAB=\angle PDC\) | Each 90⁰
\(\triangle PAB\cong \triangle PDC\) | ASA  congruence rule
AB = DC | C.P.C.T
AB = CD | C.P.C.T
(ii) Also, PB = PC
P is the mid-point of BC.

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

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)

Let, \(\angle\)DAC be \(\angle\)3
\(\angle\)1= \(\angle\)2
\(\angle\)1 + \(\angle\)3 = \(\angle\)2 + \(\angle\)3
\(\angle\)BAC = \(\angle\)EAD    ...(i)
Given that, BD = CE
BD + DC = CE + DC
\(\Rightarrow\) BC = DE      ....(ii)
\(\angle\)B = \(\angle\)E       (Given)....(iii)
From (i), (ii) and (iii), we get
\(\triangle ABC\cong \triangle AED\) (By AAS rule).

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.

AC as ㄥB is the smallest.

DE, EF

Two circles of the same radii are congruent

Two circles of the same radii are congruent

In \(\triangle\) ABC, as sum of two sides is greater than the 3rd side,
AB + BC>AC ...(1)
In \(\triangle\)ACD, as sum of two sides of a triangle is
greater than the 3rd side,
CD+ DA>AC     ....(2)
Adding (1) and (2), we get
CD + DA + AB.+ BC > 2AC.

Construction: Produce QS to meet PR in
In \(\triangle\)PQT, PQ + PT > QT
\(\Rightarrow\) PQ + PT > QS + ST       ...(1)
It \(\triangle\)SRT, TR + ST > SR      .....(2)
Adding (1) and (2), we get
PQ + PT + TR + ST > QS + ST + SR
\(\Rightarrow\) PQ + PR > QS + SR
\(\Rightarrow\) QS+SR

Since from a point \({ \bot }^{ r }\) line is the shortest. 
CF\({ \bot }\) AB
\(\therefore\) CF < AC and CF < BC ...(1) 
Similarly, BCis a line segment and A does not lie on it. AD \({ \bot }\) BC
\(\therefore\) AD < AB and AD < AC ...(2) 
Also, AC a line segment and B does not lie on it.
BE\({ \bot }\)AC
\(\therefore\) BE < AB and BE < BC ...(3)
Adding (I), (2) and (3), we get
2(AD + BE + CF) < 2(AB + BC + CA)
\(\therefore\) AB + BC + CA > AD + BE + CF 
i.e., Perimeter is greater than the sum of three altitudes. Proved.

AB = AC [Given]
 \(\therefore\) \(\angle\)ABC = \(\angle\)ACB ...(i)
BE and CF are the bisector of \(\angle\)B and \(\angle\)C
\(\angle\)EBC = \(\frac{1}{2}\)\(\angle\)ABC =\(\frac{1}{2}\) \(\angle\)ACB 
= \(\angle\)FCB
\(\angle\)EBC = \(\angle\)FCB        ....(ii)
In \(\triangle\)BEC and \(\triangle\)CEB
\(\angle\)ABC = \(\angle\)ACB
\(\angle\)EBC = \(\angle\)FCB
BC = BC
\(\therefore\) \(\triangle BEC\cong \triangle CFB\)
Hence \(\triangle EBC\cong \triangle FCB\)

In \(\triangle\)ABC 
AB = BC
\(\angle\)ABC = \(\angle\)ACB = 8
\(\angle\)B = \(\angle\)C = 8
\(\angle\)A + \(\angle\)B + \(\angle\)C = 180⁰
50° + 8 + 8 = 180⁰
28 = 180° - 50°
= 130°
8 = 65°
\(\angle\)B = \(\angle\)C = 65°
\(\therefore\) LMII BC 1
\(\therefore\) \(\angle\)LMC + \(\angle\)BCM = 180°
[\(\therefore\) \(\angle\)LMC and \(\angle\)BCM]
\(\angle\)LMC + 65° = 180°
\(\angle\)LMC = 180° - 65°
\(\angle\)LMC = 115°

\(\angle\)1 + \(\angle\)5 = 180⁰ = \(\angle\)2 + \(\angle\)6
(linear pair)
\(\Rightarrow\) \(\angle\)1 =\(\angle\)2 (\(\because\) \(\angle\)5 = \(\angle\)6)
In \(\triangle\)CAP and \(\triangle\)BAP,
\(\angle\)1 = \(\angle\)2 (proved)
\(\angle\)3 = \(\angle\)4
(AD is the bisector of \(\angle\)BAC)
AP =AP
\(\triangle CAP\cong \triangle BAP\) (By SAS) 
\(\Rightarrow\) CP = BP (By c.p.c.t.) Proved