In rectangle ABCD, P is the mid-point of AB.
Q is the mid-point of BC. R is the mid-point of CD
S is the mid-point of DA. AC is the diagonal
Now in ΔABC,
PQ = \(\frac { 1 }{ 2 } \)AC and PQ || AC        .......(1)
Similarly in ΔACD,
SR = \(\frac { 1 }{ 2 } \) AC and SR || AC        .......(2)
From (1) and (2) we get,
PQ = SR and PQ II SR
Similarly by joining BD, we have
PS = QR and PS II QR
i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.
∴ PQRS is a parallelogram.

(i) In ΔAPB and ΔCQD we have
ㄥAPB = ㄥCQD (90^o each)
AB = CD (opposite sides of parallelogram ABCD)
ㄥABP = ㄥCDQ (AB II CD and AD is a transversal)
Using ASA congruency we have,
ㄥAPB ≅ ㄥCQD
(ii) Since ΔAPB ≅ ㄥCQD
∴ Their corresponding parts are equal.
∴ AP = CQ.

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
Sum of all the angles of quadrilateral = 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac { { 360 }^{ 0 } }{ 30 } \)
=12⁰
3x =3 \(\times\) 12 = 36°
5x = 5 \(\times\) 12 = 60°
9x = 9 \(\times\) 12 = 108°
13x = 13 \(\times\) 12 = 156°
The required angles of quadrilateral are 36°, 60° 108° and 156°.

ㄥACD = ㄥA + ㄥB
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x⁰
x⁰ = 120° - 50°
= 70°
In the the triangle ABC
ㄥA + ㄥB + ㄥACB = 180° (Sum of the angles of a Δ)
50° +x + ㄥACB = 180°
50° + 70° + ㄥACB = 180°
ㄥACB = 180° - 120°
y = 60°    (OR)
ㄥACD + ㄥACB = 180⁰(Angles of a linear pair)
ㄥACB = 180° - 120°
= 60°
The value ofx = 70° andy = 60°.

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.

MPN = 90^o(Angle subtended by the diameter is 90°)
\(\angle\)OMP +\(\angle\) OPM = 180^o-120^o = 60^o
\(\angle\)MPO = \(\cfrac { 60^{ o } }{ 2 } \) = 30^o
\(\therefore\)x = \(\angle \)OPN = 90^o - 30^o = 60^o

x^o=\(\cfrac { 1 }{ 2 } \angle BOC\)
=\(\cfrac { 1 }{ 2 } \) (50^o+30^o)
=\(\cfrac { 1 }{ 2 } \) \(\times\) 80 =40^o

The distance between the two chord
FE = OF + OE
OE = \(\sqrt { { 5 }^{ 2 }-3^{ 2 } } =\sqrt { 25-9 } \)
= \(\sqrt { 16 } \) = 4 cm
OF = \(\sqrt { { 5 }^{ 2 }-{ 4 }^{ 2 } } =\sqrt { 25-16 } =\sqrt { 9 } \) = 3 cm
\(\therefore\) DE = 4cm + 3cm = 7cm
Distance between the chord is 7 cm.

Distance of the chord from the centre
= \(\sqrt { { 15 }^{ 2 }-{ 12 }^{ 2 } } \)
= \(\sqrt { 225-144 } \)
= \(\sqrt { 89 } \)
= 9 cm

ㄥP = ㄥR= 168⁰

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 \(\triangle\)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.

In an equilateral \(\triangle \) on side = 6 cm,

Construction :
Step 1: Draw ΔABC with equal sides of 6 cm length.
Step 2: Draw perpendicular bisectors of any two sides (BC and AC) to find the mid points of BC and AC.
Step 3: Draw medians BD and CE. Let them meet at G.
Step 4: Draw angle bisectors of any two sides (ㄥB and ㄥC): Let them meet at I.
Step 5: G and I are concurrent point and G is the centroid, I is the incentre.
Step 6: In an equilateral triangle centroid and incentre lie on the same point.