In figure, a diameter AB of a circle bisects a chord PQ. If AQ || PB, prove that the chord PQ is also a diameter of the circle.


Given: In figure, a diameter AB of a circle bisects a chord PQ. AQ || PB.
To Prove: The chord PQ is also a diameter of the circle.
Proof: ∠AQP = ∠ABP    ...(1)
| Angles in the same segment
∵ AQ || PB and QP intersects them
∴ ∠AQP = ∠QPB    ...(2)
| Alt. Int. ∠s
From (1) and (2),
∠ABP = ∠QPB
⇒    ∠OBP = ∠OPB
∴ OP = OB    ...(3)
| Sides opp. to equal angles
Again,
⇒ ∠BPQ = ∠BAQ    ...(4)
| Angles in the same segment
∵ AQ || PB
and AB intersects them
∴ ∠BPQ = ∠PQA    ...(5)
| Alt. Int. ∠s
From (4) and (5),
∠BAQ = ∠PQA
⇒ ∠OAQ = ∠OQA
∴ OQ = OA    ...(6)
| Sides opp. to equal angles
Adding (3) and (6), we get
OP + OQ = OB + OA
⇒ PQ = AB
∵ AB is a diameter of the circle
∴ PQ is also a diameter of the circle.

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ABCD is a cyclic quadrilateral with AD || BC. Prove that AB = DC.

Given: ABCD is a cyclic quadrilateral with AD || BC.
To Prove: AB = DC.


Given: ABCD is a cyclic quadrilateral with AD || BC.To Prove: AB = DC

Construction: Join AC.
Proof: ∵ AD || BC
and AC intersects them
∴ ∠ACB = ∠CAD    | Alt. Int. ∠s
∴ arc AB ≅ arc CD
| Arcs opposite to equal angles are congruent
∴ chord AB = chord CD
| If two arcs of a circle are congruent, then their corresponding chords are equal
⇒ AB = CD
⇒ AB = DC.
Aliter:
Given: ABCD is a cyclic quadrilateral with AD || BC.
To Prove: AB = DC.


Given: ABCD is a cyclic quadrilateral with AD || BC.To Prove: AB = DC


Construction: Draw DE || AB.
Proof: AD || BC    | Given
⇒ AD || BE
AB || DE    | By const.
∴ Quadrilateral ADEB is a parallelogram
∴ AB = DE    ...(1)
| Opp. sides of a || gm are equal
and    ∠BAD = ∠BED    ...(2)
| Opp. ∠s of a || gm are equal
∵ ABCD is a cyclic quadrilateral
∴ ∠BAD + ∠BCD = 180°    ...(3)
| ∵ Opposite angles of a cyclic quadrilateral are supplementary
∠BED + ∠CED = 180° | Linear pair
⇒ ∠BAD + ∠CED = 180°    ...(4)
| From (2)
From (3) and (4),
∠BCD = ∠CED
⇒    ∠ECD = ∠CED
∴ DE = DC
| Sides opp. to equal angles
⇒    AB = DC.    | From(1)

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ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium. 

Given: ABCD is a cyclic trapezium with AD || BC. ∠B = 70°.
To determine: Other three angles of the trapezium.
Determination:
∠B + ∠D = 180°
| ∵ Opposite angles of a cyclic quadrilateral are supplementary


Given: ABCD is a cyclic trapezium with AD || BC. ∠B = 70°.To deter

⇒ 70° + ∠D =180°
⇒    ∠D = 180° - 70°
⇒    ∠D = 110°
Again, ∵ AD || BC and transversal AB intersects them
∵ ∠A + ∠B = 180°
| ∵    The sum of the consecutive interior angles on the same side of a transversal is 180°
⇒ ∠A + 70° = 180°
⇒    ∠A = 180° - 70°
⇒    ∠A = 110°
Also, ∠A + ∠C = 180°
| ∵    Opposite angles of a cyclic quadrilateral are supplementary
⇒ 110° + ∠C = 180°
⇒    ∠C = 180° - 110°
⇒    ∠C = 70°.

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Prove that the opposite angles of an isosceles trapezium are supplementary.

Given: ABCD is trapezium in which AD = BC
To prove: Opposite angles of ABCD are supplementary.
Construction: Draw BE || AD
Proof: ∵ Quadrilateral ABED is a parallelogram


Given: ABCD is trapezium in which AD = BCTo prove: Opposite angles of

AB || DC (Given) AD || BE (By construction).
| A quadrilateral is a parallelogram if its any pair of opposite sides are parallel and equal.
∴ ∠BAD = ∠BED
| Opposite angles of a parallelogram are equal
and    AD = BE
| Opposite sides of a parallelogram are equal
But AD = BC    | Given
∴ BE = BC
∴ ∠BEC = ∠BCE
| Angles opposite to equal sides of a triangle are equal
∴ ∠BEC + ∠BED = 180° | Linear pair
⇒ ∠BCE + ∠BED = 180°
⇒ ∠BAD + ∠BED = 180°
⇒ ∠ADE + ∠ABE = 180°
⇒ Opposite angles of ABCD are supplementary.

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AC and BD are chords of a circle which bisect each other. Prove that:
(i) AC and BD are diameters.
(ii) ABCD is a rectangle.


Given: AC and BD are chords of a circle which bisect each other at O. (say).
To Prove: (i) AC and BD are diameters
(ii) ABCD is a rectangle


Given: AC and BD are chords of a circle which bisect each other at O.

Construction: Join AB, BC, CD, and DA.
Proof: (i) In ∆OAB and ∆OCD,
OA = OC
| ∵ O is the mid-point of AC
∠AOB = ∠COD
| Vertically opposite angles
OB = OD
| ∵ O is the mid-point of BD
∴ ∆OAB ≅ ∆OCD
| SAS congruence rule
∴ AB = CD    | C.P.C.T
rightwards double arrow space space space space space space space space AB with overbrace on top space approximately equal to space CD with overbrace on top space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

| If two chords of a circle are equal, then their corresponding arcs are congruent
In ∆OAD and ∆OCB,
OA = OC
| ∵ O is the mid-point of AC
∠AOD = ∠COB
| Vertically opposite angles
OD = OB
| ∵ O is the mid-point of BD
∴ ∆OAD = ∆OCB
| SAS congruence rule
∴ AD = CB    | C.P.C.T.

rightwards double arrow space space space space space space AD with overbrace on top space equals space CB with overbrace on top space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

| If two chords of a circle are equal, then their corresponding arcs are congruent
From (1) and (2),

AB with overbrace on top plus AD with overbrace on top equals CD with overbrace on top plus CB with overbrace on top
rightwards double arrow space space space space space space straight D AB with overbrace on top space equals space straight D CB with overbrace on top

⇒ BD divides the circle into two equal parts.
⇒ BD is a diameter.
Similarly, we can show that AC is a diameter.
(ii) ABCD is a parallelogram
| ∵ AB = DC and AD = BC (A quadrilateral is a parallelogram if both the pairs of opposite sides are equal)
∠ADB = 90°
| Angle in a semi-circle is 90°
∴ ABCD is a rectangle
| A parallelogram with one of its angles 90° is a rectangle




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