Chapter-10, CIRCLES

Explanation:

(D) 8 cm

If AD is 34 cm and AB is 30 cm, then

Take a look at the illustration below.

Print OL AB.

Since the chord is divided in half by the perpendicular from a circle's center to it, AL=LB=12 AB=15 cm

Right-angled ΔOLA:

OA2=OL2+AL2

∴    (17)2=OL2+(15)2

++⇒    OL2=289−225=64

∴     OL=8cm

[taking a positive square root because length is usually positive]; (17)2=OL2+(15);

As a result, the chord's distance from the middle is 8 cm.

Explanation:

(D) 8 cm

If AD is 34 cm and AB is 30 cm, then

Take a look at the illustration below.

Print OL AB.

Since a chord is divided in half by the perpendicular from a circle's center,

∴     AL=LB=12 and AB=15 cm.

In the right-angled form of OLA, 

OA2=OL2+AL2

∴    (17)2=OL2+(15)2

⇒    OL2=289−225=64

∴     OL=8cm

[taking the square root of a positive number, as length is always positive]

therefore the distance The right answer is B 2 cm.

We are aware that a chord is divided in half by the perpendicular extending from a circle's center.

8 cm = AC=CB=12AB=12; 

assuming OA = 5 cm.

Using OAC's Pythagoras Theorem,

AO2=AC2+OC2.

i.e., 52=42+OC2

=⟹OC=3 cm

[using the square root of length, which is always positive]

The radii of the same circle's OA and OD are equal.

OD is also 5 cm.

⟹CD = OD-OC = 5-3 = 2 cmThe chord's distance from the center is 8 cm.

Explanation:

The response is C.

Assuming that AB = 12 cm and BC = 16 cm

If BC and AB are in a circle, then AC will be the circle's diameter.

[circular's diameter forms a right angle to the circle]

Apply the Pythagorean theorem to the ABC right angle.

AC2=AB2+BC2⇒

AC2=(12)2+(16)2⇒

AC2=144+256⇒ AC2=400

⇒ AC2=√400=20 cm

[using the square root of a positive number since the diameter is always positive]

∴ Circle radius = 1/2(AC)=1/220=10 cm

As a result, the circle's radius is 10 cm.

Explanation:

Assumed: ABC = 20°...(1)

We are aware that an arc's angle at the circle's center is twice as large as its angle at the rest of the circle.

=><AOC = 2(20°) (From (1))

=><AOC = 40°

AOC thus equals 40 degrees.

Explanation:

The ideal selection is C.  

We are aware that the circle's diameter forms a right angle.

∠BCA = 90∘ ......(i)

AC = BC, and (ii) <ABC=<CAB (the angle opposite to equal sides is equal)

In △ ABC, ∠CBA+∠ABC+∠BCA=180∘ [Triangle's angle-sum attribute]]
∠CAB+∠CAB+90∘=180∘
2 ∠CAB=90∘
∠CAB=45∘.

Explanation:

The right answer is B 50.

Keep in mind that OA = OB = circle's radius.

<OAB=<OBA=40∘ since angles across from equal sides are also equal.

When we apply the angle sum property to △OAB, we get <AOB=180∘(40∘+40∘)=100∘.

We obtain <ACB=<AOB2=50∘ because an arc's angle at the center is twice as large as its angle at any other region of the circle.

Explanation:

Given: (1) <DAB = 60°

, (2)< ABD = 50°

 ∆ADB = 180° 110° ADB = 70°...(3) 

In ADB, DAB + DBA + ADB = 180° (angle sum property) 

=>60° + 50° + ADB = 180° (from (1) and (2))

We are aware that angles within the same circle segment are equal.

The result is that< ADB = ACB 70° 

= <ACB (From (3))<  ACB = 70°

Consequently, <ACB = 70°.

Explanation:

The right response is (B): 50

Given that the quadrilateral ABCD is cyclic and that ADC=140

We are aware that the sum of the opposite angles in a cyclic quadrilateral is 180∘, ADC+ABC=180∘, 140+ABC=180∘, ABC=180∘140∘, and ABC=40∘.

Considering that <ACB is an angle that lies within a semicircle.

The angle in a semicircle is 90∘ degrees or a right angle.

According to the angle sum property of a triangle, in ABC, <BAC+<ACB+<ABC=180∘

=>< BAC+90∘+40∘=180∘

and=>< BAC=180∘+130∘=50∘.

Explanation:

Given: BC is a circumference of a circle with a BAO of 60 degrees...

In the equation OAB, OA = OB, OAB = OBA, and OBA = 60° (angles across from equal sides are equal).

AOC also equals 60° + 60° (from (2)) and 120°...(3) (external angle) and 60° + 60° (from (3)).

We are aware that an arc's angle at the circle's center is twice as large as its angle at the rest of the circle.

As a result, <ADC = 60° and

 <AOC = 2< ADC 

=>120° = 2< ADC (from (3)).

Therefore, <ADC = 60°.

Explanation:

Given: ABC = 30° (1) and AOB = 90° (2)

We are aware that an arc's angle at the circle's center is twice as large as its angle at the rest of the circle.

As a result, <AOB = 2 ACB

=>90° = 2< ACB 

<ACB = 45°...(3)

In the case of ACB,

<CAB + <CBA + <ACB = 180° (angle sum property)

=><CAB + 30° + 45° = 180° (from (2) and (3))

<CAB + 75°= 180°75°

<CAB = 105°...(4)

Additionally, in the equation ∆OAB, OA = OB and <OAB = <OBA (angles opposed to equal sides are equal).

<OAB + <OBA +< AOB now equal 180° (angle sum property) 2OAB + 90° equals 180° (from (1) and (5)) 2<OAB = 180° 90° 2OAB = 90°

OAB = 45°...(6)

105° 45° (from (4) and (6)) = 60° is the result of <CAO =< CAB <OAB.

Therefore, CAO equals 60 degrees.

Explanation:

The right response is A True.

The above assertion is accurate since chords that are equally spaced from a circle's center have equal lengths.

AB=CD

Explanation:

Two chords are used: AB and AC.

Join us, OB and OC.

Specifically, triangle OAC and OAB

OA (common side) = OA

Triangles OAB and OAC are not congruent, therefore it is impossible to prove that any third side or angle is equivalent. OC = OB (radius of the circle).

<OAC and <OAB

The claim is untrue as a result.

Explanation:

OA = OB = O'A = O'B (Radii of the congruent circles) Two congruent circles with centers O and O' intersect at A and B. <AO′B=50o

OAOB has a rhombus shape.

The opposite angles of a rhombus are equal to one another, or <AOB=<AO′B=50o.

At this point, arc AB subtends the center of the circle, <AOB, and the remainder of the circle, <APB.

∴∠APB=12∠AOB=12×50o=25o.

Explanation:

The right response is B False.

Because a circle can travel through two unique points but not through three collinear ones, the assertion is untrue.

Explanation:

The right response is A True.

True

Consider that the diameter of a circle is AB = 6 cm.

The circle's radius is hence AB2=62=3 cm, which is correct.

Explanation:

Any circle's circumference forms a right angle with every point on the circle.

If C is a point on the circle and AOB is its diameter, then ACB is right-angled at C.

Using Pythagoras' Theorem, right-angled ACB has the formula Any circle's circumference forms a right angle with every point on the circle.

If C is a point on the circle and AOB is its diameter, then ACB is right-angled at C.

Using Pythagoras' Theorem, right-angled ACB has the formula AC2+BC2=AB2.

Explanation:

Given that ABCD is a cyclic quadrilateral with angles of 90°, 70°, 95°, and 105°,

We are aware that the sum of the opposite angles in a cyclic quadrilateral is 180°.

It can be expressed mathematically as follows: <A +< C = 90° + 95° = 185° 180°

<B +< D = 70° + 105° = 175° 180°

The sum of the opposing angles in this situation is not 180°.

It is therefore not a cyclic quadrilateral.

The claim is untrue as a result.

Explanation:

False

Because there are numerous places D where BDC=60, none of which can be the center of the circle formed by points A, B, and C.

Explanation:

True

Given that "<BAC=45° and "<BDC=45°

As we are aware, the angles in a circle's identical section are equal.

A, B, C, and D are hence concyclic.

Explanation:

Given: <ADC = 120° and AOB is a circle's diameter.

Quadrilateral ADCB is a quadrilateral that cycles.

The sum of the opposite angles in a cyclic quadrilateral is 180°.

As a result,< ADC +< CBA = 180° 

=>120° + <CBA = 180°

=><CBA = 180° 120°

=>< CBA = 180° 

=><CBA = 60°...(1).

We are aware that any point on the circle forms a right angle with the diameter.

∴ ∠ACB = 90° ...(2)

In the equation <ACB, <CAB + <CBA + <ACB = 180° (angle sum property)

<CAB + 60° + 90° = 180° (from (1) and (2)) <CAB = 180° 150° <CAB = 30°...(3)

Consequently, <CAB = 30°.

Explanation:

The right response is B 1: 1.

We possess AXBCYD. Because their corresponding chords are equivalent when two arcs of a circle are congruent, we obtain chord AB = chord CD Therefore AB: CD 1:

Explanation:

Let AB represent the chord of a circle with the origin at O. Let PQ represent the perpendicular bisector of the chord AB, which always passes through O and crosses AB at M.

In order to demonstrate this, arc PXA arc PYB Construction: Join AO and BO.

Proof In APM and BPM MB = AM (PM bisects AB)

PM is the perpendicular bisector of AB, hence <PMA =< PMB.

PM is the common side of PM.

By the SAS congruence rule, APM ≅BPM

PA = PB (using CPCT)

Consequently, Arc PXA ≅Arc PYB.

Explanation:

Given: A, B, and C are non-collinear points on a circle.

cite image

To demonstrate that AB, BC, and CA's perpendicular bisectors are concurrent.

Build together with AB, BC, and CA

Draw a perpendicular bisector between ST of AB and PM of BC of CA, where 

The ST, PM, and QR will intersect because points A, B, and C are not collinear. 

Proof: The bisector of AB and 0 lies on ST. 

∴OA=OB...(1) 

Similarly. Zero falsehoods on PM. the BC "bisector" 

∴OB=OC...(2) 

Additionally, 0 is located on QR, the CA bisector. 

∴OC=0A...(3) 

From (1). (2) and (3), 

Draw a circle C(0,r) with 0 as the center and r as the radius that will pass through the points A, B, and C. 

This demonstrates that a circle is passing. by way of points A, B, and C. Due to the fact that ST.PM and QR can only be cut at a single point zero. 

The only point that is equally far from A, B, and C is 0 (zero). 

As a result, AB.BC. and CA's perpendicular bisectors are contemporaneous. 

Explanation:

Given are two equal chords with O as their center: AB and AC.

to demonstrate that center O is located on the <BAC bisector.

Construction: Join BC and divide AD into< BAC quadrants.

Proof In ΔBAO and CAO AO = AO [same side] <BAO=<CAO

AB = AC

[According to the SAS congruence rule] ΔBAO

=>BO = CO

and <BOA = <COA are the results of the CPCT.

BO = CO, which means <BOA = <COA = 90∘.

As a result, AD is the chord BC's perpendicular bisector.

Therefore, the center O is traversed by the bisector of <BAC, or AD.

Explanation:

Given that AB and CD are two chords of a circle with O as its center, PQ is a diameter that cuts through O while also slicing apart AB and CD at L and M, respectively.

demonstrating that AB || CD

Proof that L is the chord's midpoint: OL⊥AB (because the line connecting a circle's center and its midpoint is perpendicular to the chord)

⇒ ∠ALO=90∘....(i)

Similar to that, OMCD OMD=90....(ii)

Eqs. (i) and (ii) result in ALO=<OMD=90∘.

But these are different perspectives.

So AB || CD, proving that.

Explanation:

Construction Step 1: Join CA and BD.

Step 2: Demonstrating that <CBD+<CDB=12<BAD

Having said that,

An arc's angle at the center of a circle is twice as large as its angle at any other point along the circle's circumference.

The circle's remaining portion, point B, and the center, <DAC, are subtended by the arc DC.

So,

——(1) <DAC=2 <CBD ——

At the circle's center and point D, respectively, the arc BC subtends <CAB and <CDB.

Therefore, if you combine equations (1) and (2), you get <DAC+<CAB=2<CDB+2<CBD<BAD=2(<CDB+<CBD)=>(<CDB+<CBD)=½ (<BAD).

Therefore, we demonstrate that <CBD+<CDB=½ <BAD.

Explanation:

We know that ABC is a triangle.

O is the triangle's circumcenter D is the base's midpoint BC

We must demonstrate that ∠BOD = ∠A.

Get with OB, OC, and OD.

Taking into account the triangles BOD and COD, OB = OC = circle's radius.

D being where BC's middle point is

BD = DC

OD = common side

According to the Side-Side-Side congruence rule, triangles are congruent if all three of their sides are equal to the three of another triangle.

The triangles BOD and COD are comparable using the SSS criterion.

According to the Corresponding Parts of Congruent Triangles Theorem (CPCTC), if two triangles are similar, then their corresponding sides and angles are similarly comparable or equal in size.

Using CPCTC, ∠BOD = ∠COD

∠BOC = 2 ∠BOD, then. ---------------------- (1) We are aware that an arc's angle of sustenance at the center of a circle is double its angle of sustenance elsewhere.

by it at the circle's last intersection.

∠BOC =2 ∠BAC.

Assuming (1), 2∠BOD = 2∠BAC

Consequently, ∠BOD = ∠BAC.

Explanation: 

Given, the right triangles ACB and ADB can be used to prove that <BAC=<BDC by demonstrating that <C +< D = 90° + 90° = 180°.

The fact that <BAC and <BDC are formed by the same arc in BC proves that ADBC is a cyclic quadrilateral.

Explanation: 

Given that a circle's two chords, AB and AC, intersect at the center at angles of 90° and 150°, respectively.

On the opposing sides of the middle are AB and AC.

We must locate <BAC.

Triangle BOA's OA = OB's radius when taken into consideration.

We are aware that the angles that face the same-sized sides are also equal.

Thus, <OAB = <OBA. ----------------------- (1)

OBA + OAB + AOB = 180° according to the triangle's angle sum property.

According to (1), <OAB + <OAB + <AOB = 180°

Knowing that <AOB = 90°

90° + <OAB + <OAB equals 180°.

2∠OAB = 180° - 90°

<OAB = 45° and 90° for two.

Triangle AOC: AO = OC = circle's radius

We are aware that the angles that face the same-sized sides are also equal.

Thus, <OCA = <OAC. --------------------------- (2) Given that <AOC = 150° and that 150° + <OAC +< OCA = 180°, the angle sum property of a triangle states that <AOC + <OAC + <OCA = 180°.

<OAC + <OAC = 180° - 150° according to (2).

2 OAC = 15 OAC = 30

As a result, <BAC = <OAB + <OAC = 45° + 15° = 60°.

Thus, <BAC = 60 degrees.

Explanation: 

Given that ABC is a triangle with sides AC and AB, BM and CN are its perpendiculars.

We must demonstrate the con cyclicity of B, C, M, and N.

We know that if a line segment connecting two places subtends identical angles at two other points situated on the same side of the line containing the line segment, then the four points are concyclic because BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC.

The line segment BC that connects the two points B and C in this instance subtends equal 90° angles at M and N.

The four points B, C, M, and N are hence concyclic.

Explanation: 

Given An isosceles triangle with the properties AB = AC and DE || BC is ABC.

To establish that BC is quadrilateral DE is a quadrilateral that cycles.

Construction: Draw a circle that encircles B, C, D, and E.

Proof

In ΔABC, the sides of an isosceles triangle, AB and AC, are equal.

. (i) [angles opposite to the equal sides are equal].....

Since DE || BC <ADE=<ACB, we may add EDC on both sides in Eq. (ii), giving us <ADE+<EDC=<ACB+<EDC 

=>180∘=<ACB+<EDC

=> <EDC+<ABC=180∘ [from Eq. (i)].

In light of the fact that the total of the opposite angles is 180∘, BCDE is a cyclic quadrilateral.

Explanation: 

Think about the ABCD cyclic quadrilateral.

A cyclic quadrilateral has two opposite sides that are equal.

Assuming, AD = BC

We know that the same segment of the triangle AOD and BOC subtends an equal angle to the circle (<OAD = <OBC).

Assuming AD = BC, ODA = OCB

According to the ASA criterion, two triangles are congruent if any two of their included sides and related angles are equivalent in size to those of the other triangle.

The triangles AOD and BOC are comparable using the ASA criterion.

<AOD and <BOC

Triangle DOC added on both sides, <AOD + <DOC +< BOC + <DOC +< ADC + <DOC + <BCD

According to the Corresponding Parts of Congruent Triangles Theorem (CPCTC), if two triangles are similar, then their corresponding sides and angles must also be comparable.

are also comparable, or have the same dimensions.

AC = BD using CPCTC.

Therefore, it is established that if two of the diagonals of a cyclic quadrilateral are equal, then so are its opposite sides.

Explanation: 

Given, ABC is a triangle

O is the circumcentre of the triangle

We have to prove that ∠OBC + ∠BAC = 90º.

Become a part of BO and CO.

Let OBC = OCB equal. --------------------------- (1) Taking triangle OBC into account

In accordance with the angle sum property, <BOC +< OCB +< CBO = 180°; <BOC + θ+θ  = 180°; and <BOC = 180 - 2θ°. ---------------------------------- (2) We are aware that an arc's angle at the center of a circle is twice as large as its angle at the rest of the circle.

As a result, <BOC = 2 <BAC <BAC = 1/2 <BOC From (2), <BAC = 180° - 2θ°/2 <BAC = 90° - θFrom (1), <BAC + <OBC = 90°

<OBC plus <BAC equals 90º as a result.

Explanation: 

Given that AB is a circle's chord and that it equals the circle's radius, join OA, AC, and BC.

AB = OA=OB because OA = OB = radius of the circle.

OAB is an equilateral triangle as a result.

⇒ ∠AOB=60∘ [An equilateral triangle has 60° for each angle]

According to the theorem, an arc's angle at the center of a circle is twice as large as its angle at the rest of the circle.

Specifically, <AOB=2 <ACB" <ACB=60∘" "2=30∘"

At one point in the major portion, the chord AB subtends an angle of 30°.

Explanation:

Assumed: <ADC = 130∘ (1) chord BC = chord BE (2)

Quadrilateral ADCB is a quadrilateral that cycles.

The sum of the opposite angles in a cyclic quadrilateral is 180°.

Consequently,< ADC +< CBA = 180°

=>130° +< CBA = 180°

=><CBA = 180°

=><CBA = 50°...(3).

BC equals BE (given) in CBO and EBO

OB = OB (frequent)

OC is equal to OE (circular radius).

Through SSS property,  ∆OCB ≅ ∆OEB

Consequently, according to C.P.C.T., <OBC = <OBE = 50°...(4)

As a result, <CBE = <OBE + <OBC 

= 50° + 50° (From (4))

= 100°.

Therefore, CBE = 100°.

Explanation: 

Assumed, ACB = 40^o

We need to locate <OAB.

We are aware that an arc's angle at the center of a circle is twice as large as its angle at the rest of the circle.

<ACB = 2 <AOB = 2

(40º) <AOB = 80º ----------------------- (1) Taking triangle AOB into consideration, OA = OB = circle's radius

We are aware that the angles that face the same-sized sides are also equal.

<OAB =< OBA ------------------- (2) 

From (1) and (2), 80º + <OAB + <OAB = 180º according to the triangle's angle sum property: <AOB + <OBA + <OAB = 180º

2∠OAB = 180º - 80º

<OAB = 100º/2 and 2<OAB = 100º

<OAB thus equals 50º.

Explanation:

Given that <ADC=130∘," draw a quadrilateral ABCD enclosed in a circle with the letter O as its center.

ABCD becomes a cyclic quadrilateral because it is a quadrilateral that is encircled by a circle.

A cyclic quadrilateral's opposite angles added together equal 180∘, as shown by the formulas ADC+ABC=180∘, 130+ABC=180∘, and ABC=50∘.

Given that AB is a circle's diameter, AB's angle with the circle is a right angle.

∴ ∠ACB=90∘

With respect to <ABC, <BAC+<ACB+<ABC=180∘ [by the angle sum property of a triangle] <BAC+90∘+50∘=180∘ <BAC=180∘(90∘+50∘) <BAC=180∘140∘=40∘

Explanation:

Given: At locations A and B, two circles with centers O and O′ intersect.

Draw a line parallel to OO′ through B, perpendicular to PQ at OX and O′Y, and unite them all.

We are aware that the chord can be divided by a perpendicular taken from its center.

∴PX=XB, YQ=BY, 

and PX+YQ=XB+BY [on combining the two equations above]

PX+YQ+XB+BY=2(XB+BY) 

PX+BX+BY+YQ=2(XB+BY)

PQ=2(XY) 

∴PQ=2(OO′) [∵XY=OO′] are the results of adding XB+BY on both sides.

Thus, PQ=2OO′

Explanation: 

Join AE in the given figure.

ACDEE is a cyclic quadrilateral because it contains four points on a circle, A, C, D, and E.

The sum of the opposite angles in a cyclic quadrilateral, <ACD+<AED, is 180∘.

Now that AEB=90∘, the diameter forms a right angle to the circle in (ii).

The result of combining Equations (i) and (ii) is (<ACD+<AED)+(<AEB=180∘+90∘)=270∘, where <ACD+<BED=270∘.

Explanation: 

Assuming <OAB=30∘ and <OCB=57∘

The radius of the same circle is the same in the expression AO = OB.

Angles across from sides with the same length are equal.

=> <AOB+<OBA+<BAO=180∘ [by the triangle's angle sum property]

<AOB+30∘+30∘=180∘

 <AOB=180∘(2(30∘) 

<AOB=180∘(60∘)=120∘ (i)

Now, in the equation OCB, OC = OB (both represent the radius of a circle).

[Angles opposite to equal sides are equal] <OBC = <OCB = 57∘

According to the triangle's angle sum property, <COB+<OCB+<CBO=180∘.

∠COB=180∘−(∠OCB+∠OBC)

 ∠COB=180∘−(57∘+57∘) 

∠COB=180−114∘=66∘

According to Eq. (i), the following values are obtained: <AOB=120∘, <AOC+COB=120∘

< AOC+66∘=120∘

 <AOC=120∘

66∘=54∘.

Explanation: 

The two equal chords should be AB and CD. AB = CD. Let point E be where the chords converge. Adopt OE.

AE = CE and BE = DE must be proven.

From the center O to the chords, draw perpendiculars. The chords AB at M and CD at N are split in half by this perpendicular.

As a result, AM = MB = CN = DN.(1) For ONE and OME, <M = <N = 90°, OE = OE, and OM = ON. The distance between equal chords and the center is equal.

According to RHS standards, OME and ONE are congruent.

Me = NE according to CPCT, thus. (2) We are aware that AE = AM + ME and CE = CN + NE.

It is clear from (1) and (2) that CE equals AE.

BE = AB - AE, DE = CD - CE, and

Equals include AB, CD, CE, and AE. therefore, DE and BE are equally so. It has been established that chords with comparable segments are equal.

Explanation:

We are aware that a quadrilateral is cyclic if the sum of any two of its opposite angles is 180°. 

Draw an ABCD trapezium using AB || CD.

The non-parallel sides that are equal are AD and BC. AD = BC. BN ⊥ CD and AM⊥   CD are drawn.

Think about  ΔAMD and  ΔBNC.

(Given) AD = BC

<AMD = <BNC (90 degrees)

AM = BN (The angle at which two parallel lines are perpendicular is the same)

AMD BNC, by RHS congruence.

<ADC equals <BCD using CPCT, and <ADC and <BAD are on the same side of transversal AD.

<ADC +< BCD + <BAD equals 180 degrees. [From formula (1)]

This equation demonstrates the complementary nature of the sum of opposite angles. ABCD is a cyclic quadrilateral as a result.

Explanation: 

Given, ABC is a triangle

P, Q, and R are the midpoints of the sides BC, CA, and AB

The perpendicular to A on BC is AD.

We need to demonstrate the con cyclicity of P, Q, R, and D.

Rejoin DR, RQ, and QP.

Think about a right-angled triangle. The midpoint of AB is ADP, R.

Thus, RB = RD.

We are aware that equal sides have equal-sized angles opposing them.

∠2 = ∠1 --------------------------------- (1)

The midpoint theorem asserts that "The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half the length of the third side" because R and Q are the midpoints of AB and AC.

The midpoint theorem states that RQ || BC RQ || BP.

We are aware that a parallelogram's opposite sides are parallel and congruent.

Since RB, BPQ, QPR is a parallelograms, and we are aware that a parallelogram's opposite angles are equal.

∠1 = ∠3 ------------------------------- (2)

2 equals 3 from (1) and (2) ------------------------ (3) The linear pair of angles is equivalent to 180 degrees, as we are aware.

∠2 + ∠4 = 180

From (3), ∠3 + ∠4 = 180

We are aware that a cyclic quadrilateral's pair of opposite angles equal 180 degrees.

The points P, Q, R, and D are hence concyclic.

Explanation: 

ABCD is a parallelogram, therefore

Draw a circle across points A and B so that it crosses AD at points P and Q.

We must demonstrate the con cyclicity of P, Q, C, and D.

Participate in PQ

By a cyclic quadrilateral's exterior angle property, 1 = A

We are aware that a parallelogram's opposing angles are equal since <A = <C.

So, ∠1 = ∠C --------------------------- (1)

The total of the internal angles on the same side of the transversal, AD || BC and PQ, is 180 degrees.

So, ∠C+ ∠D = 180°

We can deduce from (1) that the total of the opposite angles in a cyclic quadrilateral is 180 degrees.

As a result, the quadrilateral QCDP cycle.

The points P, Q, C, and D are hence con-cyclic.

Explanation: 

Think of the ABC triangle.

We must demonstrate that the circumference of ABC is where the perpendicular bisector of BC and the angle bisector of A intersect.

Let P be the point where the angle bisector of A intersects the ABC circumcircle.

Get with BP and CP.

We are aware that the angles within a single circle segment are equal.

Thus, <BAP =< BCP.

Given that AP divides <A in half.

<BCP = <BAP

<A = <BAP plus <BCP.

<BAP = <BCP = 1/2< A, therefore. -------------------------- (1)

Likewise, <PAC = <PBC = 1/2 <A. ------------------- (2)

<BCP = <PBC from (1) and (2)

We are aware that if the angles that two chords of a circle's center subtend are equal, then the chords themselves are also equal.

As a result, P is on the perpendicular bisector of BC and BP = CP.

Consequently, the perpendicular bisector and angle bisector of A

BC and ABC cross at the center of ABC.

Explanation: 

Assume that two chords AB and CD meet in a circle at a right angle. AYDZBWCX

The statement "arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle" must be demonstrated.

Create a circle with a diameter EF and center M.

Since EC = DF and CD||EF arc EC --------------------- (1) The arcs ECXA and EWB and AF and BF are equal. ---------------------- (2)

Arc ECXAYDF = semicircle, as we are aware.

So, arc EA plus arc AF equals a semicircle.

Arc EC + Arc CXA + Arc BF = Semicircle from (2)

Arc DF + Arc CXA + Arc BF = Semicircle from (1)

semicircle = arc DF + arc BF + arc CXA

The figure shows that arc DF + arc BF = arc DZB.

Arc DZB plus Arc CXA equals a semicircle.

Since the circle divides into two semicircles, we know that the remaining part of the circle is also a semicircle.

So, semicircle = arc AYD + arc BWC.

Explanation: 

Given that P can be any point on the minor arc BC that does not overlap with B or C, ABC is an equilateral triangle inscribed in a circle.

To demonstrate that PA is a bisector of an angle

Building: Integrate PB and PC

ABC is an equilateral triangle, as shown by the proof that 3+4=60∘.(i) Right now, the angles 1=4=60∘ [in the same segment AB] and <2=<3=60∘ [in the same segment AC]

From (i), we have (1+2+60∘).

PA is hence the bisector of< BPC.

Explanation: 

Construct: Sign up for AD

In order to demonstrate: <AEC=1/2 (angles occupied by arcs CXA and DYB at the center).

We are aware that an arc's angle at the center of a circle is twice as large as its angle on the rest of the circle.

In this case, arc AXC subtends the circle's center (<AOC) and point D (<ADC).

∴∠AOC=2∠ADC ⇒∠ADC=1/2∠AOC...(1)

Additionally, arc <DYB subtends angles <DOB at the circle's center and <DAB at point A.

∴∠DOB=2∠DAB ⇒∠DAB=1/2∠DOB...(2)

Now, in the equation <ADE, we have <AEC=<ADC+<DAB (Exterior angle) <AEC=1/2<AOC+<DOB [<DAB=<DOB, angle subtended by an identical arc at different places on the circle are equal]

Consequently, <AEC=1/2 (angle subtended by arc CXA at the center + angle subtended by eqs. (1) and (2)by the central arc DYB).

Explanation: 

Build along with QC and QD.

A cyclic quadrilateral is ABCD.

The total of the opposing angles in a cyclic quadrilateral, <CDA+<CBA, is 180∘.

The result of multiplying both sides by 2 is 1/2<CDA+1/2<CBA=1/2x180∘=90∘, where =><1+<2=90∘. [<1=1/2 CDA and< 2=1/2 <CBA] (i)

However, [angles in the same QC section are equal] <2=<3 ….(ii)

Eqs. (i) and (ii) result in PDQ=90∘.

PQ is therefore a circle's diameter since the circle's diameter subtends a right angle at the circumference.

Explanation: 

Create a circle with O in the center. The chord of a circle is defined as AB = 2 cm. The line OM separates the chord AB into two equally spaced halves.

In order to demonstrate that AN = NB = 1 cm and OB = √2 cm,

On the basis of ONB, OB2=ON2+NB2 [Pythagorean theorem].

ON2=21=1 ON = 1 cm [getting positive square root, since the distance is always positive] => (√2)2=ON2+(1)2

=> ON2=21=1 ON = 1 cm

The angles opposed to equal sides are also equal, and ONB=90∘ [ON is the perpendicular bisector of the chord AB]; NOB=NBO=45 [ON =NB =1 cm];

Likewise, <AON=45∘

Therefore, <AOB=<AON+<NOB =45∘+45∘=90∘.

We are aware that the angle a chord subtends to a circle is equal to half of the angle it subtends to the center.

It was proven because <APB=12<  AOB <APB=90∘2=45∘.

Explanation: 

Given a circle with two equal but intriguing chords (AB and CD), point P

to demonstrate PB=PD.

Join OP and draw OL⊥AB, and CD for construction.

Proof: The chords AB=CD OL=OM are equally spaced from the center.

We have the equations OL = OM [established above] and OLP=OMP [each 90].

Additionally, according to the RHS congruence criterion, OP=OP [common side] <OLP=<OMP

[by CPCT] LP=MP ……(i)

Now, divide both sides by two to get AB=CD

=> 1/2(AB)=1/2(CD).

BL=DM...............(ii) [perpendicular line drawn perpendicularly from circle's center to circle bisects chord, i.e., AL=LB and CM=MD]

The result of subtracting equation (ii) from equation (i) is LP-BL=MP-DM PB=PD. thus proved.

Explanation:

Two chords AB and AC exist in a circle of radius r such that AB = 2AC. Additionally, AB and AC are separated from the center by p and q, respectively.

demonstrating 4q2=p2+3r2

Proof

Let AB = 2a if AC = a.

The chords AC and AB are drawn perpendicularly from the center O at the points M and N, respectively. AM=MC=a2 AN = NB = a

In the formula OAM, AO2=AM2+MO2 [by Pythagoras theorem] and AO2=(a2)2+q2. (i)

According to the Pythagorean theorem, AO2=a2+p2 in the equation OAN. (ii)

From Eqs. (i) and (ii), (a2)2+q2=a2+p2 (a2)2+2

=a24+q2=a2+p2 (a2)2+4q224a2+2 [multiplying both sides by 4] (a2)2+4q2=3a2+4p2 (a2)2+4q2

=p2+3(a2+p [R2=a2+p2 in right-angled OAN]

So it was proved.

Explanation: 

SUMMARY: In the above diagram, OD and BC are parallel.

<BCO = <COD (Alternate Interior Angles), where <COD = 30°.(1) We are aware that an arc's angle at the centre of a circle is twice as large as its angle on the rest of the circle.

In this instance, arc CD subtends the circle's centre <COD and point B <CBD).

<COD = 2 <CBD = 30°/2 = 15°

y = 15° (from (1))

...(2) In addition, the circle's arc AD subtends the points AOD in the centre and ABD at B.

<AOD = 2 <ABD, where <ABD=90°/2=45°.(3) In ∆ ABE,

 x + y + ∠ ABD + ∠ AEB = 180 ∘

 ⇒ x + 15 ∘ + 45 ∘ + 90 ∘ = 180 ∘

 ⇒ x = 180 ∘ − (90 ∘ + 15 ∘ + 45 ∘ ) 

⇒ x = 180 ∘ − 150 ∘ ⇒ x = 30 ∘ (Sum of the angles of a triangle) (from (2) and (3))

Thus, x = 30°.

Explanation: 

Given that in the diagram CD AB and BD = OD

In the formula OBD, 

BD = OD and OD = OB (both the radius of the same circle).

BD = OB + OD

ODB is an equilateral triangle as a result.

'BOD=' 'OBD=' 'ODB=60'

MB = MB [common] in MBC and MBD

∠CMB=∠BMD=90∘

Also, CM = MD Any perpendicular drawn on a chord in a circle will also cut the chord in half.

∴ΔMBC and MDB [by SAS congruence]

∴<MBC = <MBD [by CPCT]

∴<MBC=' '<OBD=60∘

Since AB is a circle's diameter, ACB=90∘.

In the equation ACB<CAB+<CBA+<ACB=180∘ [by the triangle's angle sum condition], thus CAB=180∘(60∘+90∘)=30∘.