Given: ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively.
To Prove: AX IICY.
Proof: \(\because\) ABCD is a parallelogram.
\(\therefore\) \(\angle \)A = \(\angle \)C  I Opposite \(\angle \)s of a parallelogram are equal

⇒  \(1\over 2\)\(\angle \)A=\(1\over 2\)\(\angle \)C     |\(\because \)Halves of equals are equal
⇒    \(\angle \)1 = \(\angle \)2    ...(1)      I    \(\because\)   AX is the bisector of \(\angle \)A and CY is the bisector of \(\angle \)C Now, AB II DC and CY intersects them
 \(\therefore\)  \(\angle \)2 =\(\angle \)3  ....(2)   I Alternate interior \(\angle \) s 
From (1) and (2), we get
\(\angle \)1 = \(\angle \)3
But these form a pair of equal corresponding angles
\(\therefore\)  AX II CY.

Given: In a quadrilateral ABCD, the line segments bisecting \(\angle \)C and \(\angle \)D meet at E.
  To Prove:   \(​​\angle \)A+\(​​\angle \)B = 2 \(\angle \)CED
I Angle sum property of a quadrilateral
In \(\Delta \) CED,
\(\angle \)CED +\(\angle \)EDC+\(\angle \)ECD = \(180°\)   | Angle sum property of a triangle
⇒ \(\angle \)CED+\(1\over2\)\(\angle \)D+\(1\over2\)\(\angle \)C=\(180°\)
 ⇒ 2\(\angle \)CED +\(\angle \) D+\(\angle \)C =\(360°\)    ...(2)
From (1) .and (2),
2\(\angle \)CED+\(\angle \)D+C=A+B+C+D
 ⇒ 2 \(\angle \)CED =\(\angle \)A+\(\angle \)B
 ⇒ \(\angle \)A+\(\angle \)B=2\(\angle \)CED

Given:    ABCD is a parallelogram. AC is a diagonal of parallelogram ABCD which divides it into two triangles, namely, \(\Delta \) ABC and \(\Delta \)CDA

To Prove: \(\Delta \)ABC  \(\cong \) \(\Delta \)CDA
Proof: BC IIDA  IOpposite sides of a parallelogram are parallel and    AC   is a transversal
 \(\angle \)  BCA=\(\angle \)DAC  ...(1)  I Pair of alternate interior angles  
 Also, AB IIDC     IOpposite sides of a parallelogram are parallel and AC is a transversal
\(\therefore \) \(​​\angle \)BAC = \(\angle \)DCA   ......(2)   I Alternate interior angles      
AC = CA      ...(3)    I Common  
In view of (1), (2) and (3),
\(\Delta \) ABC \(\cong \) \(\Delta \) CDA      I ASA congruence criterion

Let us recall what a rectangle is.
A rectangle is a parallelogram in which one angle is a right angle.
 
Let ABCD be a rectangle in which \(\angle\) A = 90°.
We have to show that \(\angle\) B = Ð C = \(\angle\) D = 90°
We have, AD || BC and AB is a transversal
(see Fig.).
So, \(\angle\) A + \(\angle\) B = 180° (Interior angles on the same
side of the transversal)
But, \(\angle\) A = 90°
So, \(\angle\) B = 180° – \(\angle\) A = 180° – 90° = 90°
Now, \(\angle\) C = Ð A and \(\angle\) D = \(\angle\) B
(Opposite angles of the parallellogram)
So, \(\angle\) C = 90° and \(\angle\) D = 90°.
Therefore, each of the angles of a rectangle is a right angle.

AM丄DC, AN丄BC
In quadrilateral AMCN,
ㄥA+ㄥM+ㄥC+ㄥN=360°
ㄥA+ㄥC=180°
⇒ 50°+ㄥC=180° ⇒ ㄥC=130°

In parallelogram, ㄥA=ㄥC=130°
ㄥB=ㄥD=180°-130°
=50°

ㄥADC+ㄥBCD=180°
⇒ \(\frac{1}{2}\)ㄥADC+\(\frac{1}{2}\)ㄥBCD=90°
or ㄥ1+ㄥ2=90°

In ΔODC,
ㄥ1+ㄥ2+ㄥDOC=180°
ㄥDOC=90°

∠APR=ㄥDRP
or ㄥ1=ㄥ2
But these are alternate interior angles
SP II RQ, SR II PQ
PQRS is a parallelogram
∠APR+ㄥBPR=180°,(linear pair)
⇒ \(\frac{1}{2}\) ∠APR+\(\frac{1}{2}\)ㄥBPR=\(\frac{1}{2}\)x180°
⇒ ∠1+ㄥ3=90°
⇒ ∠SPQ=90°

PQRS is a rectangle

Let the measure of the angles be x, 2.x, 3x and 4x then,
x + 2x + 3x + 4x = 360°
⇒ x = 36°
ஃ Angles of quadrilateral are 36°, 72°, 108°,144°

Since opposite angles of a parallelogram are equal
3x - 2 = 63 - 2x
⇒ x=13°
Angles of a parallelogram :
(39 - 2)°, (180- 37)°, (63 - 26)°, (180- 37)°
i.e., 37°,143°,37°,143°.

Sum of the angles of a quadrilateral is 360°.
ஃ 4x° + 7x° + 15x° + 10x° = 360°
[Angle sum property of quadrilateral]
⇒ 36x°= 360°
⇒ x=10°
ஃ Smallest angle = 4x° = 40°
Largest angle = 15x°= 150°

10.5 cm

72°, 108°, 72°, 108°

\(36°\),\(144°\),\(36°\), \(144°\)

40°, 80°, 100°, 140°; Trapezium

\(\angle \)OAB+\(\angle \)OBA=\(\frac { 1 }{ 2 } \)\(​​\angle \)A+\(\frac { 1 }{ 2 } \)\(\angle \)B
=\(\frac { 1 }{ 2 } \)(\(\angle \)A+\(\angle \)B)=(\(180°\))=\(90°\)
\(\therefore \) \(\angle \)AOB=\(90°\)

x+\(80°\)=3x- \(1 0°\)\(\Rightarrow \)   x=\(45°\)

\(\angle \)AGF +\(\angle \) GAE = \(​​180°\)
\(\Rightarrow \) \(\angle \)AGF+\(​​60°\)=\(​​180°\)
\(\Rightarrow \)\(\angle \)AGF=\(​​120°\)

See a rhombus

See a blackboard

no of angle of a rhombus is 90⁰