\(\angle\)FEH =\(\angle\)AED = 180⁰-\(\left( \frac { 1 }{ 2 } \angle A+\frac { 1 }{ 2 } \angle B \right) \)

\(\angle\)FGH =\(\angle\)CGB
= 180⁰-\(\left( \frac { 1 }{ 2 } \angle A+\frac { 1 }{ 2 } \angle B \right) \)
Adding,
\(\angle\)FEH +\(\angle\)FGH = 180⁰
\(\therefore\)  EFGH is a cyclic quadrilateral.

Given: AB and CD are equal chords intersecting at 90°
To prove: OPQR is a square
Proof: Since P and R are the mid-point of AB and CD respectively
\(\therefore\) \(\angle\) OPB = \(\angle\)ORD = 90°
\(\Rightarrow\)\(\angle\)OPQ = \(\angle\)ORQ = 90° 
Since equal chords on a circle are equidistant from the centre.
\(\therefore\) OP= OR
Thus in t10PQ and t10RQ, we have
OP=OR
\(\angle\)OPQ= \(\angle\)ORQ
and OQ= OQ
\(\therefore\) \(\triangle OPB\cong \triangle ORQ\)
Thus in quadrilateral OPQR,
We have
OP = OR, PQ = RQ
and \(\angle\)OPQ = \(\angle\)ORQ = 90°
Hence OPQR is a square.

Draw: OM \(\bot\) AB and ON.\(\bot\) CD.
To Prove: AB = CD

In \(\Delta\)OME and \(\Delta\)ONE,
\(\angle\)OEM = \(\angle\)OEN          (given)
\(\angle\)OME = \(\angle\)ONE          (each 90^o)
OE= OE                        (common)
\(\therefore\) By A.A.S. congruence rule
\(\Delta\)OME: \(\Delta\)ONE
\(\therefore\) OM = ON
Hence, AB = CD (Chords equidistant from the centre are equal)

Given: ABC is a triangle and P is a point on the side BC such that AB = AP. AP produced meets the circumcircle of \(\Delta ABC\) at Q.
To Prove: CP=CQ

Proof: In \(\Delta ABP\) and \(\Delta CQP\),
 \(\angle BAP=\angle QCP\)
| Angles in the same segment of a circle are equal
\(\angle ABP=\angle CPQ\)
| Vertically opposite angles
∴ \(\Delta ABP\cong \Delta CPQ\)
| AA criterion of similarity
∴ \(\frac { AB }{ CQ } =\frac { BP }{ QP } =\frac { AP }{ CP } \)
| ∵ Corresponding sides of two similar triangles are proportional
⇒ ​​​​​​\(\frac { AB }{ CQ } =\frac { AP }{ CP } \)
But  AB=AP   | Given
∴ CQ=CP

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: \(\angle AQP=\angle ABP\)   ...........(1)
   I Angles in the same segment of a circle are equal
∵ AQ || PB and QP intersects them
∴ \(\angle AQP=\angle QPB\)   ....(2)
| Alt. Int. \(\angle s\)
From (1) and (2),
\(\angle ABP=\angle QPB\)
⇒ \(\angle OBP=\angle OPB\)
∴ OP=OB                ...(3)
| Sides opposite to equal angles of a triangle are equal
Again,
\(\angle BPQ=\angle BAQ\)           ...(4)
I Angles in the same segment of a circle are equal
∵ AQ || PB
and AB intersects them
\(\angle BPQ=\angle PQA\)            ....(5)
| Alt. Int. \(\angle s\)
From (4) and (5),
\(\angle BAQ=\angle PQA\)
⇒ ​​​\(\angle OAQ=\angle OQA\)
∴ OQ=OA                    ....(6)
I Sides opp. to equal angles of a triangle are equal
Adding (3) and (6), we get
OP + OQ = OB + OA
PQ = AB
∵ AB is the diameter of the circle
∵ PQ is also the diameter of the circle.

Given: ABCD is trapezium in which AD=BC
To prove: Opposite angles of ABCD are supplementary.
Construction: Draw BE || AD
Proof: In quadrilateral ABED,
AB || DE  I Given
AD || BE  | By construction
∴ Quadrilateral ABED is a parallelogram.

| A quadrilateral is a parallelogram if its both the pairs of opposite sides are parallel.
∴ \(\angle BAD=\angle BED\)  ..........(1)
I Opposite angles of a parallelogram are equal
But  AD=BC   | Given
∴ BE=BC
∴ \(\angle BEC=\angle BCE\) | Angles opposite to equal sides of a triangle are equal
∴ \(\angle BEC=\angle BED=180°\) | Linear pair axiom
 ​​​​​​​\(\Rightarrow \angle BCE+\angle BED=180°\)   | From (2)
 \(\Rightarrow \angle BCE+\angle BAD=180°\)   | From (1)
\(\Rightarrow \angle BCD+\angle BAD=180°\)
\(\Rightarrow \) Opposite angles of ABCD are supplementary.

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

Construction: Join AC.
Proof: ∵  AD || BC and AC intersects them
∴ ​​​​​​ \(\angle ACB=\angle CAD\)    I Alt. Int. L s
∴ ​​​​​​​\(arc\quad AB\cong arc\quad CD\)
| Arcs corresponding 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.

Given: ABCD is a cyclic quadrilateral whose diagonals intersect at P. \(\angle DBC=70°\) and \(\angle BAC=30°\).
Required: To find \(\angle BCD\) .
Determination: \(\angle BDC=\angle BAC(=30°)\)
| Angles in the same segment of a circle are equal
Now, in \(\Delta BCD\) ,
\(\angle BCD+\angle BDC+\angle DBC=180°\)
| ∵ The sum of the three angles of a \(\Delta \) is 180°
⇒  \(\angle BCD+30°+70°=180°\)
⇒  \(\angle BCD+100°=180°\)
⇒  \(\angle BCD=180°-100°=80°\).

Given: AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E.
To Prove: EB = ED and AE = CE.

Construction: From O draw OP丄 AB and OQ丄CD. Join OE.
Proof: AB = CD
∴  OP=OQ I ∵ Equal chords of a circle are equidistant from the centre
Now in right \(\quad \Delta s\) OPE and OQE,
Hyp. OE = Hyp. OE  I Common
Side OP = Side OQ  I Proved above
∴ \(\Delta OPE\cong \Delta OQE\)
I R.H.S. Congruence Axiom
∴  PE= QE      I CPCT
⇒ \(PE-\frac { 1 }{ 2 } AB=QE-\frac { 1 }{ 2 } CD\)
| ∵ AB = CD (Given)
⇒ PE-PB = QE-QD
⇒ EB = ED.
⇒ BE + AB = ED + CD I ∵ AB = CD     | ∵ AB = CD
⇒ AE=CE

Given: A circle with centre O. PQ is a chord of this circle.
OL is the perpendicular drawn to chord PQ from centre O.
To Prove: PL = QL
Construction: Join OP and OQ.

Proof: In \(\Delta OLP\) and 
OP=OQ             I Radii of the same circle
OL = OL            I Common
\(\angle OLP=\angle OLQ\)     I Each = 90°
∴   \(\Delta OLP\cong \Delta OLQ\)    I By RHS congruence criterion
∴ PL=QL      ICPCT