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)
\({ x }^{ 0 }+{ x }^{ 0 }+{ x }^{ 0 }={ 180 }^{ 0 }\)
\(3{ x }^{ 0 }={ 180 }^{ 0 }\)
\({ x }^{ 0 }={ 60 }^{ 0 }\)
\(\angle A=\angle B=\angle C={ 60 }^{ 0 }\)

Given: l and m are two parallel lines intersected by another pair of parallel lines p and q
To Prove: \(\triangle ABC\cong \triangle CDA\)
Proof: \(AB\parallel DC\) and \(AD\parallel BC\)
Quadrilateral ABCD is a parallelogram.
| A quadrilateral is a parallelogram if both the pairs of opposite sides are parallel
BC = AD ........... (1) | Opposite sides of a \(\parallel \) gm are equal
AB = CD .............(2) | Opposite sides of a \(\parallel \) gm are equal
\( \angle ABC=\angle CDA\)......(3)  | Opposite angles of a \(\parallel \) gm are equal
In  \(\triangle ABC\) and \(\triangle CDA\),
AB = CD | From (2)
BC = DA | From (1)
\( \angle ABC=\angle CDA\)  | From (3)
\(\triangle ABC\cong \triangle CDA\) | SAS Rule

Given ABCD is a quadrilateral in which AD=BC and \(\angle DAB=\angle CBA\) 
To prove: (i) \(\triangle ABD\cong \triangle BAC\)
(ii) BD = AC
(iii) \(\angle ABD=\angle BAC\)
Proof: (i) In \(\triangle ABD\) and \(\triangle BAC\)
AD = BC
AB = BA
\(\angle DAB=\angle CBA\)
\(\triangle ABD\cong \triangle BAC\)  |SAS Rule
(ii) \(\triangle ADB\cong \triangle BAC\)  |Proved in (i)
BD = AC  | C.P.C.T
(iii) \(\triangle ABD\cong \triangle BAC\) |Proved in (i)
\(\angle ABD=\angle BAC\)  | C.P.C.T

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: 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)

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.

In \(\triangle\)ADC and \(\triangle\)ABC,
AC is Common
\(\angle DAC=\angle BAC\)
\(\angle DCA=\angle BCA\)(Given)
Hence, \(\triangle ADC\cong \triangle ABC\) (By AAS rule)
\(\Rightarrow\) CD = BC (By c.p.ct) Proved

AC as ㄥB is the smallest.

BC, AB

Theorem

Theorem

Theorem

Obviously AB=PQ

Two circles of the same radii are congruent

↔ denotes correspondence

≌ represents congruence.

\(\angle\)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