Proof: In \(\triangle\)AOB,
\(\angle\)B >\(\angle\)A
\(\angle\)A <\(\angle\)B
OB In \(\triangle\)COD,
\(\angle\)C >\(\angle\)D
\(\angle\)D < \(\angle\)C 
OC < OD      ....(2)
Adding (1) + (2),
OB + OC < OA + OD
BC or, AD> BC

PQ = PR \(\Rightarrow\)  \(\angle\)PQR = \(\angle\)PRQ
(Angles opp. to equal sides) 
ST II QR \(\Rightarrow\) \(\angle\)PST = \(\angle\)PQR
(Corresponding angles)
\(\Rightarrow\) PTS = \(\Rightarrow\)PRQ
(Corresponding angles) 
\(\therefore\) PST = PTS 
\(\Rightarrow\) PS = PT.
(Sides opp. to equal angles)

AB =AC
\(\therefore\) \(\angle\)B = \(\angle\)C = x
BA =BC
\(\therefore\) \(\angle\)A = \(\angle\)C = x
AC =BC
\(\angle\)A = \(\angle\)B = x
But, \(\angle\)A + \(\angle\)B + \(\angle\)C = 180°
Also,
3x = 1800
\(\angle\)A = \(\angle\)B = \(\angle\)C = 60°.
Alternative Method:

Let \(\triangle\)ABC be an equilateral triangle, so that AB = AC = BC
Now, AB = AC
\(\Rightarrow\) \(\angle\)B = \(\angle\)C ....(1)
(\(\because\)Angles opp. to equal sides are equal)
CB = CA
\(\Rightarrow\) \(\angle\)A = \(\angle\)B ....(2)
(\(\because\) Angles opp. to equal sides are equal)
From (1) and (2), we have
\(\angle\)A = \(\angle\)B = \(\angle\)C
Also, \(\angle\)A +\(\angle\).B +\(\angle\)C = 180°,(Angle sum property)
\(\therefore\) \(\angle\)A + \(\angle\)A + \(\angle\)A = 180°
\(\Rightarrow\) 3\(\angle\)A = 180° \(\Rightarrow\) \(\angle\)A = 60°.
\(\therefore\) \(\angle\)A = \(\angle\)B = \(\angle\)C = 60°.
Thus, each angle of an equilateral triangle is 60°.

AB = BC
\(\Rightarrow\) \(\frac{1}{2}\)AB = \(\frac{1}{2}\) BC
But \(\frac{1}{2}\) AB = BX
\(\frac{1}{2}\) BC = BX
But
\(\frac{1}{2}\) BC = BY
From (i) and (ii), BX = BY

BR + BQ = CQ + BQ \(\Rightarrow\) QR = BC
\(\triangle ABC\cong \triangle PQR\) by RHS\(\cong\)condn
\(\therefore\) AC = PR (By c.p.c.t.)

(a) 60⁰
(b) 60⁰

40⁰

Draw the perpendicular bisectors of AB and AC. Their point of intersection is the required point.

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: 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