\(\angle\)In the cyclic quadrilateral \(\angle\) A+\(\angle\)C = 180^o
y + 4^o + 3y^o + 8^o = 180^o
4y^o + 12^o = 80^o
4y^o = 180^o -12^o = 168
y =\(\cfrac { 168 }{ 4 } \) = 42
\(\angle\)B + \(\angle\)D = 180^o
8x + 12 = 180^o
8x = 180^o - 12^o = 168
x = \(\cfrac { 168 }{ 8 } \) = 21^o
\(\therefore\) \(\angle\)A = y + 4^o = 42^o + 4^o = 46^o
\(\angle\)C = 3y + 8 = 3 \(\times\) 42 + 8 = 126 + 8 = 134^o
\(\angle\)B = 3x + 6 = 3 \(\times\) 21 + 6 = 63 + 6 = 69^o
\(\angle\)D = 5x + 6 = 5 \(\times\) 21 + 6
= 105 + 6 = 111^o

We have a parallelogram ABCD in which diagonals AC bisect ㄥA.
ㄥDAC = ㄥBAC
(i) To prove that AC bisects LC
∵ ABCD is a parallelogram
∴ AB II DC and AC is a transversal
∴ ㄥ1 = ㄥ3 (Alternate interior angle)     .........(1)
.Also BC II AD and AC is a transversal.
∴  ㄥ2 = ㄥ4 (Alternate interior angle)  ...........(2)
But AC bisects ㄥA
∴ ㄥ1 = ㄥ2
From (1), (2) and (3) we get
ㄥ3 = ㄥ4
∴ AC bisects ㄥC.
(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ㄥ1 = ㄥ4 [ ∵ ㄥ1 = ㄥ2 = ㄥ4]
∴ BC = AB (side opposite to equal angles are equal) (4)
Similarly AD = DC            ....... (5)
But ABCD is a parallelogram AB = DC (Opposite sides of a parallelogram)        .......(6)
From (4), (5) and (6) we have AB = BC = CD = DA.
Thus ABCD is a rhombus.

A parallelogram in which bisector of angle A, B, C, D intersect at P, Q, R, S to form a quadrilateral PQRS.
To prove: Quadrilateral PQRS is a rectangle.
Proof: Since ABCD is a parallelogram. Therefore, AB II DC.
Now, AB II DC, and transversal AD cuts them, so we have
\(\angle A+\angle D=180^0\)
\(\frac{1}{2}\angle A+\frac{1}{2}\angle D=\frac{180^0}{2}\)
\(\angle DAS+\angle ADS=90^0\)
But in \(\triangle ASD\) , we have
\(\angle ADS+\angle DAS+\angle ASD=180^0\)
\(90^0+\angle ASD=180^0\)
\(\angle ASD=90^0\)
\(\angle RSP=\angle ASD\) (vertically opposite angle)
\(\angle RSP=90^0\)
Similarly, we can prove that
\(\angle SRQ=90^0,\angle RQP=90^0\ and \angle QPS=90^0\)
Thus, PQRSis a quadrilateral each of whose angle is 90°.
Hence, PQRS is a rectangle.

\(\angle A=72^0\)
\(\angle C=180^0-72^0\) (\(\angle A\) and \(\angle C\) are supplementary)
= 108⁰
\(\angle A+\angle B+\angle C+\angle D=360^0\) (Total angles of a quadrilateral)
72^o+ 2x -10 +108^o+ x + 4 = 360^o
3x + 174^o = 360^o
3x = 360^o-174^o = 186⁰
\(x=\frac{186^0}{3}\)
x = 62^o
The value of x = 62°
\(\angle B=2x-10\)
= 2(62^o) -10
= 124^o-10^o = 114^o
\(\angle D=x+4\)
= 62^o+ 4 = 66^o
The other angles are 72°, 114°, 108° and 66°.

Given ∠BAD = 120° and AC bisects ∠BAD
\(\angle{BAC}=\frac{1}{2}\times 120^{0}=60^{0}\)
∠1 = ∠2 = 60°
AD || BC and AC is the traversal
∠2 = ∠4 = 60°
Δ ABC is isosceles triangle    [∴ ∠1 = ∠4 = 60 °]
⇒ AB = BC
Parallelogram ABCD is a rhombus.

In \(\triangle ABC\angle B=60^0\ and\ \angle C=70^0\)
\(\therefore\angle A=180^0-(60^0+70^0)\)
\(=180^0-130^0\) = 50⁰
In \(\triangle E=70^0\ and \angle D=60^0\)
\(\therefore \angle F=180^0-(70^0+60^0)\)
= 180° - 130° = 50°
\(\angle A=\angle F=50^0\)
\(\angle B=\angle D=60^0\)
\(\angle C=\angle E=70^0\)
By AAA congruency
\(\therefore\triangle ABC\cong\triangle FDE\)

Construction:
Step 1: Draw ΔABC with the given measurements AB = 8 cm, \(\angle\)A = 90^o and AC = 6 cm and construct the perpendicular bisector of any two sides (AB and AC) to find the mid points M and N of AB and BC respectively.
Step 2: Draw the medians (C and BN and let them meet at G. The point G is the centroid of the given ΔABC.

In \(\triangle\)PQR, PQ = 5 cm, PR = 6 cm, \(\angle\)QPR = 60°

Construction:
Step 1: Draw \(\triangle\) PQR using the given measurements PQ = 9 cm, QR = 7 cm and RP = 8 cm and construct the perpendicular bisector of any two sides (PQ and QR) to find the mid-points M and N of PQ and QR respectively.
Step 2: Draw the medians PN and RM and let them meet at G. The point G is the centroid of the given \(\triangle\)PQR.

In \(\triangle\)ABC, AB = BC = 8 cm, \(\angle\)B = 70°.

Construction :
Step 1: Draw \(\triangle\)ABC with BC = 8 cm, \(\angle\)B = 70°. AB = 8.
Step 2: Construct the angle bisectors of any two angles (B and C) and let them meet at I. Then I is the incentre of \(\triangle\)ADC. Draw perpendicular from I to anyone of the side (BC) to meet BC at D.
Step 3: With I as centre and ID as radius draw a circle. This circle touches all the sides of the triangle internally.

Construction :
Step 1: Draw \(\triangle\)ABC with AB = BC = CA = 8 cm
Step 2: Construct angle bisectors of any two angles (A and B) and let them meet at I. I is the incentre of \(\triangle\)ABC.
Step 3: Draw perpendicular from I to any one of the side (AB) to meet AB at D.
Step 4: With I as centre, ID as radius draw the circle. This circle touches all the sides of triangle internally.