Let ABC be a right angled triangle in which ㄥB = 90°.
Then, ㄥA +ㄥC = 90°
|Sum of all the angles of a triangle is 180°
ㄥB=ㄥA+ㄥC
ㄥB>ㄥA and ㄥB>ㄥC
AC>BC
| Side opposite to greater angle is longer
and AC> AB

AC is the longest side, i.e., hypotenuse is the longest side.

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: In \(\triangle ABC\) , AD is the perpendicular bisector of BC
To Prove:  \(\triangle ABC\) is isosceles triangle in which AB = AC.
Proof: In \(\angle ADB=\angle ADC\)
DB = DC | AD is the perpendicular bisector of BC
AD = AD
\(\triangle ADB\cong \triangle ADC\) |By SAS rule
AB = AC  | C.P.C.T
\(\triangle ABC\)  is an isosceles triangle in which
AB = AC 

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

Let, BE\(\bot \)AC and CF\(\bot \)AB
In \(\triangle\)s ABE and ACF, we have
\(\angle\)AEB = \(\angle\)AFC (.: Each = 90°) 
\(\angle\)A = \(\angle\)A (Common)
and BE = CF (Given)
\(\therefore\) By AAS criterion at congruence, we have
\(\triangle ABE\cong \triangle ACF\)
\(\Rightarrow\) AB = AC.
(\(\therefore\) Corresponding parts of congruent triangles are equal)

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.

We have AP ⊥ BC [Given]
∴ ∠APB = 90^०and APC = 90^०
In ΔABP and ΔACP, we have
∠APB = ∠APC [each = 90^०]
AB = AC [Given]
AP = AP [Common] 
∴ Using RHS criteria. ΔABP ≅ ΔACP
∴ Their corresponding parts are congruent.
⇒ ∠B = ∠C

Only to prove that ΔANC≅ΔDMB by SAS congruence rule.

In \(\Delta\)PQR, \(\angle\) P=\(\angle\)R
PQ=QR
[sides opposite to equal angle are equal]
\(\Rightarrow \) PQ=4cm [\(\because \) QR=4cm]

DE, EF

Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively.
To Prove: BE = CF
Proof: ABC is an isosceles triangle
AB = AC
\(\angle ABC=\angle ACB\) | Angles opposite to equal sides of a triangle are equal
In \(\triangle BEC\) and \(\triangle CFB\)
\(\angle BEC=\angle CFB\)  | Each = 90⁰
BC = CB
\(\angle ECB=\angle FBC\)
\(\triangle BEC\cong \triangle CFB\) | By AAS Rule
BE = CF | C.P.C.T

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

Two squares of the same side length ar congruent

Two circles of the same radii are congruent

Two equilateral triangles of the same side length are congruent

Given ABCD is a square and ΔDEC is an equilateral triangle.
To prove (i) ΔADE ≅ ΔBCE
(ii) AE = BE
(iii) ㄥDAE = 15⁰
Proof From figure, we have
ㄥADE = ㄥADC + ㄥCDE
= 90⁰ + 60⁰ = 150⁰  ...(i)
[since, each angle of a square is 90° and each angle of an equilateral triangle is 60°]
Similarly ㄥBCE=ㄥBCD+ㄥDCE
= 90⁰ + 60⁰ = 150⁰ ...(ii)
[since, each angle of a square is 90° and each angle of an equilateral triangle is 60°]
(i) In ΔADE and ΔBCE,
AD = BC  [sides of a square]
DE = CE  [sides of an equilateral tiangle]
and ㄥADE = ㄥBCE = 150⁰  [from Eqs. (i) and (ii)]
∴  ΔADE ≅ ΔBCE  [by SAS congruence rule]
(ii) From part (i) ΔADE ≅ ΔBCE
Then, AE = BE  [by CPCT]
(iii) In ΔADE,
AD = BE  [∵ AD = DC = DE]
⇒  ㄥAED = ㄥDAE  ...(iii)
[since, angles opposite to equal sides of a triangle are equal]
Now, ㄥDAE + ㄥAED + ㄥADE = 180⁰
[by angle sum property of a triangle]
⇒ ㄥDAE + ㄥDAE + 150⁰=180⁰
[from Eqs. (i) and (iii)]
​​​​​​⇒  2ㄥDAE = 30⁰
∴ ㄥDAE = 15⁰

Given \(\triangle ABC\) is an isosceles triangle with
AB = AC
OB and OC are the bisectors of LB and LC respectively and intersect each other at O.
i.e. and To prove (i) OB = OC
(ii) AO bisects Proof (i) In MBC, we have
AB = AC [given)
\(\Rightarrow \angle B=\angle C\)
[since, angles opposite to equal sides are equal)
\(\Rightarrow \frac { 1 }{ 2 } \angle B=\frac { 1 }{ 2 } \angle C\)
\(\Rightarrow \angle OBC=\angle OCB\)
\(\angle OBA=\angle OCA\)
[since sides opposite to equal angles of a triangle are equal]
(iii) In \(\triangle ABO\ \triangle ACO\) [given]
AB = BC[given]
\(\angle OBA=\angle OCA\)
and OB = OC
\(\therefore \triangle ABO\equiv ACO\)
[by SAS congruence rule]
Then \(\angle BAO=\angle CAO\) [by CPCT[]
So AO is the bisector of \(\angle BAC\)
Hence proved

Given AC = AE, AB = AD and \(\angle BAD=\angle EAC\) 
To prove BC=DE
proof We have  \(\angle BAD=\angle EAC\) 
On adding \(\angle DAC\) to both sides, we get
\(\angle BAD+\angle DAC=\angle EAC+\angle DAC\)
\(\\ \angle BAC=\angle DAE\)
In \(\triangle ABC\quad \triangle ADE,AC=AE\)
AB = BD
and \(\angle BAC=\angle DAE\)
\(\therefore \triangle ABC\cong \triangle ADE\) [by SAS congruence rule]
Then , BC = DE [by CPCT]
Hence Proved.