In \(\Delta\) ABC and \(\Delta\) ADC
AB = AD
| One pair of consecutive sides BC = DC
| Other pairs of consecutive sides AC = AC | Common
\(\therefore\) \(\Delta\) ABC \(\cong \) \(\Delta\) ADC
| SSS Congruence Axiom

\(\therefore \angle BAC=\angle DAC\) | CPCT
\(\angle BCA=\angle DCA\) | CPCT
i.e., diagonal AC bisects \(\angle BAD\) and \(\angle BCD\) each.

The kite has only one line of symmetry shown by dotted line segment (a diagonal) AC. and \(\angle B=\angle D\)
The diagonal AC is an angle bisector as it bisects \(\angle A\) and \(\angle C\) both.
Yes; the diagonals cut at right angles.
No; the diagonals are not equal in length.
Yes; one of the diagonals bisect the other.

In \(\Delta\)COD,  we have \(\angle\)COD + \(\angle\)1 + \(\angle\)2 = 180^o
\(\Rightarrow\) \(\angle\)COD = 180^o - [\(\angle\)1 + \(\angle\)2]
\(\Rightarrow\)\(\angle\)COD = 180^o - \(\left[ \frac { 1 }{ 2 } \angle D+\frac { 1 }{ 2 } \angle C \right] \)
\(\Rightarrow\) \(\angle\)COD = 180^o - \(\frac { 1 }{ 2 } \left[ \angle D+\angle C \right] \)
But \(\angle\)A + \(\angle\)B + \(\angle\)C + \(\angle\)D = 360^o
\(\Rightarrow\) \(\angle\)C + \(\angle\)D = 360^o - (\(\angle\)A + \(\angle\)B )
\(\therefore\)\(\angle\)COD = 180^o - \(\frac { 1 }{ 2 } \)[360^o - (\(\angle\)A + \(\angle\)B)
= 180^o - \(\frac { 1 }{ 2 } \)[360^o] + \(\frac { 1 }{ 2 } \)[\(\angle\)A + \(\angle\)B]
= 180^o - 180^o + \(\frac { 1 }{ 2 } \)(\(\angle\)A + \(\angle\)B) = \(\frac { 1 }{ 2 } \)(\(\angle\)A + \(\angle\)B)
Thus, \(\angle\)COD = \(\frac { 1 }{ 2 } \)[\(\angle\)A + \(\angle\)B]

Let ABCD be a parallelogram such that \(\angle\)A and \(\angle\)B are 4x and 5x respectively.
Since, the adjacent angles are supplementary,
\(\therefore\)\(\angle\)A + \(\angle\)B = 180°
4x + 5x = 180°
\(\Rightarrow\) 9x = 180^o \(\Rightarrow\) x = \(\frac { { 180 }^{ o } }{ 9 } \)=20^o
\(\therefore\)\(\angle\)A = 4x = 4 x 20^o = 80^o and \(\angle\)B = 5x = 5 x 20^o
We know that opposite angles of a parallelogram are equal.
\(\angle\)C = \(\angle\)A = 80° And \(\angle\)D = \(\angle\)B = 100°
Thus, \(\angle\)A = 80°, \(\angle\)B = 100°, \(\angle\)C = 80° and \(\angle\)D = 100°

\(\because \)\(\angle\)A : \(\angle\)B : \(\angle\)C : \(\angle\)D = 1 : 2 : 3 : 4
\(\therefore\)Let us suppose that
\(\angle\)A = 1x^o, \(\angle\)B = 2x^o
\(\angle\)C = 3x^o, \(\angle\)D = 4x^o
Since, \(\angle\)A + \(\angle\)B + \(\angle\)C + \(\angle\)D = 360°
:. x + 2x + 3x + 4x = 360°
\(\Rightarrow\)10x = 360°\(\Rightarrow\)x = \(\frac { { 360 }^{ o } }{ 10 } \) = 36^o
\(\therefore\)Angles are: \(\angle\)A = x^o = 36°
\(\angle\)B = 2x^o = 2 x 36° = 72°
\(\angle\)C = 3x^o = 3 x 36° = 108°
\(\angle\)D= 4x^o = 4 x 36° = 144°
Thus, the measure of the angles of the quad. are
36°, 72°, 108° and 144°

Since, the sum of the measures of interior angles of a quadrilateral is 360°.

Also, 115° + 70° + 60° = 245°
\(\therefore\)245° + LABC = 360°
\(\Rightarrow\)\(\angle \)ABC = 360° - 245° = 115°
Now, x = ext. \(\angle\)BCD = 180° - \(\angle\)BCD
= 180° - 115° = 65°
Similarly, y = 180° - 70° = 110°
z = 180° - 60° = 120°
w = 180° - 115° = 65°
\(\therefore\)x +y + z + w = 65° + 110°+ 120° + 65° = 360°

128.5⁰

12 cm

72°

(a) 108°
(b) 135°

We have, FARE is a trapezium, where ER II FA and EP and RP are bisectors of LE and LR respectively.
Thus, ㄥPEF = ㄥPER and ㄥPRE = ㄥPRA
∴  ㄥPEF = 25° and ㄥPRA = 30°
So, ㄥE = 25° + 25° = 50°
and ㄥR = 30° + 30° = 60°
Now, we know that in a trapezium,
ㄥE + ㄥF =180° and ㄥR + ㄥA =180°
∴  ㄥF =180° - 50° =130°
and ㄥA =180° - 60° =120°
or ㄥEFA =130° and ㄥFAR =120°

We know that, the sum of exterior angles of a regular polygon is 360°.

So, x⁰+ (3x)⁰ + (4x + 10)⁰ + (2x + 5)⁰+ (5x)⁰ = 360⁰
⇒ 15x+15⁰ =360⁰ ⇒ 15x=360⁰-15⁰
⇒ 15x=345⁰ ⇒ \(\frac { { 345 }^{ 0 } }{ 15 } \)=23⁰
∴ 3x⁰ = 3 x 23 = 69° ,
4x + 10 = 4 x 23 + 10= 92 + 10 = 102° ,
2x + 5 = 2 x 23 + 5 = 46 + 5 = 5t~,
5x = 5 x 23 = 115°
x =23°
So, exterior angles are 23° ,69° ,102° ,51° and 115°.

Let given polygon be
ABCDE It is clear that, AB is a straight line.
90° + ㄥ1 = 180°
[by linear pair angle]
⇒ ㄥ1 = 180⁰ - 90⁰ = 90⁰
Now, x +90° +60° +90° + 70° = 360°
The sum of the exterior angles of any polygon is 360°.
⇒ x + 310⁰ = 360°
⇒ x = 360⁰-310⁰=  50°
Hence, the measure of angle x is 50°.

We know that, the sum of three angles of a triangle is 180°.

Let the given triangle be ABC.
∴ ㄥA + ㄥB+ ㄥC =180°
⇒  30⁰ + 90⁰+ ㄥC=180° ⇒  ㄥC+120⁰=180°
⇒  ㄥC = 180° - 120° = 60°
Now, ㄥEBC + ㄥABC = 180° [by linear pair]
∴  x +90° = 180° ⇒ x = 180° - 90° = 90°
Similarly, ㄥFCA + ㄥACB = 180° [by linear pair]
⇒ y+60⁰=180° ⇒ y=180⁰-60⁰=120°
and ㄥDAB + ㄥBAC = 180° [by linear pair]
⇒ z + 30° = 180° ⇒ z = 180° - 30° = 150°
∴ x + y + z = 90° + 120° + 150° = 360°
Hence, the value of x + y + z is 360°

If a quadrilateral is not convex, then it will be concave. In the quadrilateral ABCD, join BD. Then, quadrilateral split into two triangles.

In ΔADB, we have
mㄥ1 + mㄥ2 + mㄥ3 = 180°
[using angle sum property of a traingle]    ... (i)
In ΔBDC, we have
mㄥ4 + mㄥ5 + mㄥ6 = 180°
[using angle sum property of a triangle]     ... (ii)
On adding Eqs. (i) and (ii), we get
mㄥ1 + mㄥ2 + mㄥ3 + mㄥ4 + mㄥ5 +mㄥ6 = 180°+180°
⇒ mㄥ1 + mㄥ2 + mㄥ6+ mㄥ5 + (mㄥ3 + mㄥ4) = 360°
∴ mㄥA + mㄥB + mㄥC + mㄥD = 360°
Hence, the sum of the interior angles of a quadrilateral is 360°.