Important theorem on circle and

Quadrilaterals

 

what are the circle theorem in geometry ?

the circle theorems in geometry are statements that prove important result about circle. these theorems provide important information or fact about various parts of a circle.

what are the Quadrilaterals theorem in geometry ?

Quadrilaterals theorems are rules and properties that apply to quadrilaterals 4 sided polygons. These theorems help identify types of quadrilaterals and solve for angles, side, and other properties

theorem on circle

 

A perpendicular drawn from the Centre of a circle on its chord bisects the chord.

 

 

 

 

 

 

Given:- seg AB is a chord of a circle with Centre O.

Seg OP ┴ chord AB

To prove :-seg AP ≅ Seg BP

Proof  :-  Draw seg OA and seg OB

In ΔOPA and Δ OPB

∠OPA ≅ ∠OPB……..seg OP ┴ Chord AB

Seg OP ≅ seg OP ………(Common side)

Hypotenuse OA ≅ hypotenuse OB ….(radii of the same circle)

∴ Δ OPA ≅ Δ OPB …. (Hypotenuse side theorem)

Seg PA ≅ Seg PB…..(c.s.c.t)

 

The Segment joining the centre of a circle and the midpoint of its cord is perpendicular to the chord.

 

 

 

 

 

 

Given:-seg AB is a chord of a circle with centre  O and P is the midpoint of chord AB of the circle. That means seg AP ≅ seg PB.

To prove :- seg OP ┴ chord AB

Proof :- Draw seg OA and seg OB.

In Δ AOP and Δ BOP

Seg OA ≅ seg OB …..(radii of the same circle)

Seg OP ≅ seg OP ….( Common sides)

Seg AP ≅ seg BP…(Given)

∴ ΔAOP ≅ ΔBOP…..(SSS test)

∴ ∠OPA ≅ ∠OPB……(C.a.c.t)   (I)

Now ∠ OPA +∠OPB=180̊…. (Angle in line pair)

∴ ∠OPB + ∠OPB=180̊……(from I)

∴ 2 ∠OPB=180̊

∴ ∠OPB=90̊

∴ seg OP ┴ chord AB.

 

Congruent chords of a circle are equidistant from the centre of the circle.

 

 

 

 

 

Given:- In a circle with centre O.

Chord AB ≅ Chord CD

OP ┴ AB, OQ ┴ CD

To prove:- OP= OQ

Construction:- Join seg OA and seg OD.

Proof:- AP=1/2 AB, DQ=1/2 CD…(Perpendicular drawn from the centre of a circle to its chord bisects the chord.)

AB = CD…………………(Given)

∴ AP =DQ

∴Seg AP ≅ Seg DQ………(I)… segments of equal lengths.

In right angled Δ APO and right Δ DOQ

Seg AP ≅ seg DQ…….(from I)

Hypotenuse OA ≅ Hypotenuse OD …..(Radii of the same circle)

∴ Δ APO ≅ Δ DQO…….(hypotenuse side theorem)

Seg OP ≅ Seg OQ…… (c.s.c.t.)

∴ OP=OQ……….(Length of congruent segments.)

Congruent chords in a circle are equidistant form the centre of the circle.

Important theorem on Quadrilaterals

 

Opposite sides and opposite angles of a parallelogram are congruent.                                                               

 

 

 

 

Given:- □ABCD is a parallelogram.

∴  side AB║side DC, side AD ║side BC,

To prove :- seg AD ≅ seg BC;seg DC ≅ seg AB

∠ADC ≅ ∠CBA, and ∠DAB ≅∠BCD.

Construction:- Draw diagonal AC.

Proof:- seg DC║seg AB and diagonal AC is a transversal.

∴∠DCA ≅ ∠BAC……….(1)

And ∠DAC ≅ ∠BCA………(2)}……alternate angles

Now, in ΔADC and ΔCBA,

∠DAC ≅ ∠BCA….. from (2)

∠DCA ≅ ∠BAC….. from (1)

Seg AC ≅ seg CA… common side

∴ ΔADC ≅ Δ CBA…..ASA test

∴ side AD ≅ side CB……… c.s.c.t.

And side DC ≅ side AB…….c.s.c.t

Also, ∠ADC ≅ ∠CBA……… c.a.c.t

Similarly we can prove ∠DAB ≅ ∠ BCD.

 

Diagonals of  parallelogram bisect each other.

 

 

 

 

Given:- □ PQRS is a parallelogram.

Diagonals PR and QS intersect in point O.

To prove :- seg PO ≅ seg RO,

Seg SO ≅ seg QO.

Proof:- In ΔPOS and ΔROQ

∠OPS ≅ ∠ORQ…… alternate angles

Side PS ≅ side RQ…… opposite side of parallelogram

∠PSO ≅ ∠RQO ……….alternate angles

∴ ΔPOS ≅ ΔROQ……… ASA test

∴seg PO ≅ seg RO….. { corresponding side of

And seg SO ≅ seg QO…..  congruent triangles}

 

If pairs of opposite sides of a quadrilateral are congruent then that quadrilateral is a parallelogram. 

 

 

 

 

 

Given :- In □PQRS

Side PS ≅ side QR

Side PQ ≅ Side SR

To prove:- In □PQRS is as parallelogram .

Construction : Draw diagonal PR

Proof :- In Δ SPR and ΔQPR

Side PS ≅ Side QR….(Given)

Side SR ≅ Side QP…..(Given)

Side PR ≅ Side RP….(Common side)

∴ ΔSPR ≅ Δ QPR…..(SSS test)

∴ ∠SPR ≅∠QPR…..( c.a.c.t.)

Similarly, ∠PRS ≅ ∠RPQ….(c.a.c.t.)

∠SPR and ∠QPR are alternate angles formed by the transversal PR of seg PS and seg QR.

∴ side PS║ side QR …(I) alternate angles test for parallel lines.

Similarly ∠PRS and ∠RPQ are the alternate angles formed by transversal PR of seg PQ and seg SR.

∴ side PQ║side SR…..(II)…..alternate angle test

∴ from (I) and (II) □PQRS is parallelogram.

 

Diagonals of a rectangle are congruent . 

 

 

 

 

 

Given:- □ABCD is rectangle.

To prove :- Diagonal AC ≅ Diagonal BD

Proof:- Complete the proof by giving suitable reason.

Side DC ≅ Side AB{Properties rectangle .}

∠ADC ≅ ∠DAB  (Opposite sides are equal and all four angles are 90̊)

Side AD ≅ Side AD (Common side )

Δ ADC ≅ ΔDAB ……(SAS test)

∴  Diagonal AC ≅ Diagonal BD….. (c.s.c.t.)