Mix Examples - Circles Questions in English

(A) The length of an arc of a circle is given by the formula $L = \frac{\theta}{360^{\circ}} \times 2\pi r$,where $\theta$ is the angle subtended by the arc at the center and $r$ is the radius of the circle.
Since the two circles are congruent,they have the same radius,let it be $r$.
Let $L_1$ be the length of arc $AXB$ and $L_2$ be the length of arc $A'YB'$.
Given that the angle subtended by arc $AXB$ at center $O$ is $\theta_1 = 75^{\circ}$.
So,$L_1 = \frac{75^{\circ}}{360^{\circ}} \times 2\pi r$.
Given that the angle subtended by arc $A'YB'$ at center $O'$ is $\theta_2 = 25^{\circ}$.
So,$L_2 = \frac{25^{\circ}}{360^{\circ}} \times 2\pi r$.
The ratio of the lengths of the arcs is $\frac{L_1}{L_2} = \frac{\frac{75^{\circ}}{360^{\circ}} \times 2\pi r}{\frac{25^{\circ}}{360^{\circ}} \times 2\pi r} = \frac{75^{\circ}}{25^{\circ}} = \frac{3}{1}$.
Therefore,the ratio is $3:1$.

(B) In $\triangle OPQ$,since $OP$ and $OQ$ are perpendiculars from the center to equal chords,$OP = OQ$.
Therefore,$\triangle OPQ$ is an isosceles triangle,which implies $\angle OQP = \angle OPQ = k$.
In $\triangle OPQ$,the sum of angles is $180^{\circ}$,so $\angle OQP + \angle OPQ + \angle POQ = 180^{\circ}$.
$k + k + 150^{\circ} = 180^{\circ}$.
$2k = 30^{\circ}$.
$k = 15^{\circ}$.
Thus,$\angle OPQ = 15^{\circ}$.
Since $OP \perp AB$,$\angle OPA = 90^{\circ}$.
From the figure,$\angle OPA = \angle OPQ + \angle APQ$.
$90^{\circ} = 15^{\circ} + \angle APQ$.
$\angle APQ = 90^{\circ} - 15^{\circ} = 75^{\circ}$.

(C) Let $O$ be the centre of the circle. Draw $OP \perp AB$.
Since the perpendicular from the centre to a chord bisects the chord,we have:
$AP = \frac{1}{2} \times AB = \frac{1}{2} \times 30 = 15 \, cm$.
The radius of the circle is $OA = \frac{AD}{2} = \frac{34}{2} = 17 \, cm$.
In the right-angled triangle $\Delta OPA$,by the Pythagorean theorem:
$OP^2 + AP^2 = OA^2$
$OP^2 + 15^2 = 17^2$
$OP^2 + 225 = 289$
$OP^2 = 289 - 225 = 64$
$OP = \sqrt{64} = 8 \, cm$.
Thus,the distance of the chord $AB$ from the centre is $8 \, cm$.

(D) Since the perpendicular from the centre of a circle to a chord bisects the chord,we have:
$AC = \frac{1}{2} \times AB = \frac{1}{2} \times 8 \text{ cm} = 4 \text{ cm}$.
In the right-angled triangle $\triangle OCA$,by the Pythagoras theorem:
$OC^2 + AC^2 = OA^2$
$OC^2 + 4^2 = 5^2$
$OC^2 + 16 = 25$
$OC^2 = 25 - 16 = 9$
$OC = \sqrt{9} = 3 \text{ cm}$.
Since $OD$ is the radius of the circle,$OD = OA = 5 \text{ cm}$.
Therefore,$CD = OD - OC = 5 \text{ cm} - 3 \text{ cm} = 2 \text{ cm}$.

(A) $AB$ is perpendicular to $BC$,therefore $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$.
Since the triangle is right-angled,the hypotenuse $AC$ is the diameter of the circle passing through points $A, B$ and $C$.
Using the Pythagorean theorem in $\Delta ABC$:
$AC = \sqrt{AB^2 + BC^2}$
$AC = \sqrt{12^2 + 16^2}$
$AC = \sqrt{144 + 256}$
$AC = \sqrt{400} = 20 \, cm$
Since the diameter $AC = 20 \, cm$,the radius $r = \frac{AC}{2} = \frac{20}{2} = 10 \, cm$.

(B) According to the theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Here,arc $AC$ subtends $\angle AOC$ at the centre $O$ and $\angle ABC$ at point $B$ on the remaining part of the circle.
Therefore,$\angle AOC = 2 \times \angle ABC$.
Given $\angle ABC = 20^{\circ}$.
Thus,$\angle AOC = 2 \times 20^{\circ} = 40^{\circ}$.

(C) Since $AOB$ is a diameter of the circle,the angle subtended by it at the circumference is $90^{\circ}$.
Therefore,$\angle ACB = 90^{\circ}$ (Angle in a semi-circle is $90^{\circ}$).
Given that $AC = BC$,the triangle $\triangle ABC$ is an isosceles triangle.
In an isosceles triangle,angles opposite to equal sides are equal,so $\angle CAB = \angle CBA$.
Let $\angle CAB = \angle CBA = x$.
In $\triangle ABC$,the sum of angles is $180^{\circ}$:
$\angle CAB + \angle CBA + \angle ACB = 180^{\circ}$
$x + x + 90^{\circ} = 180^{\circ}$
$2x = 90^{\circ}$
$x = 45^{\circ}$
Thus,$\angle CAB = 45^{\circ}$.

(D) In $\Delta OAB$,
$OA = OB$ (Radii of the same circle).
Therefore,$\angle OAB = \angle OBA = 40^{\circ}$ (Angles opposite to equal sides are equal).
In $\Delta OAB$,the sum of angles is $180^{\circ}$.
So,$\angle AOB = 180^{\circ} - (40^{\circ} + 40^{\circ}) = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
According to the theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 100^{\circ} = 50^{\circ}$.

(A) In $\Delta ADB$,the sum of angles is $180^{\circ}$.
$\angle DAB + \angle ABD + \angle ADB = 180^{\circ}$
$60^{\circ} + 50^{\circ} + \angle ADB = 180^{\circ}$
$110^{\circ} + \angle ADB = 180^{\circ}$
$\angle ADB = 180^{\circ} - 110^{\circ} = 70^{\circ}$
Since angles in the same segment of a circle are equal,we have $\angle ACB = \angle ADB$.
Therefore,$\angle ACB = 70^{\circ}$.

(B) Given that $ABCD$ is a cyclic quadrilateral and $AB$ is the diameter of the circle.
Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
$\angle ADC + \angle ABC = 180^{\circ}$
Given $\angle ADC = 140^{\circ}$,we have:
$140^{\circ} + \angle ABC = 180^{\circ}$
$\angle ABC = 180^{\circ} - 140^{\circ} = 40^{\circ}$
Since $AB$ is the diameter,the angle subtended by the diameter at the circumference is $90^{\circ}$.
Therefore,$\angle ACB = 90^{\circ}$.
In $\Delta ABC$,the sum of angles is $180^{\circ}$:
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$
$\angle BAC + 40^{\circ} + 90^{\circ} = 180^{\circ}$
$\angle BAC + 130^{\circ} = 180^{\circ}$
$\angle BAC = 180^{\circ} - 130^{\circ} = 50^{\circ}$

(C) In $\Delta OAB$,we have:
$OA = OB$ [Radii of the same circle]
Therefore,$\angle ABO = \angle BAO$ [Angles opposite to equal sides are equal].
Given $\angle BAO = 60^{\circ}$,so $\angle ABO = 60^{\circ}$.
In $\Delta OAB$,the sum of angles is $180^{\circ}$,so $\angle AOB = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ}$.
Since $BC$ is a diameter,$\angle AOC = 180^{\circ} - \angle AOB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Thus,$\angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 120^{\circ} = 60^{\circ}$.

(D) In $\Delta OAB,$ we have $OA = OB$ (radii of the same circle).
Therefore,$\angle OAB = \angle OBA.$
Since the sum of angles in a triangle is $180^{\circ},$
$2 \angle OAB = 180^{\circ} - \angle AOB = 180^{\circ} - 90^{\circ} = 90^{\circ}.$
Thus,$\angle OAB = 45^{\circ}.$
Also,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
So,$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 90^{\circ} = 45^{\circ}.$
Now,in $\Delta CAB,$ the sum of angles is $180^{\circ}.$
$\angle CAB = 180^{\circ} - (\angle ABC + \angle ACB) = 180^{\circ} - (30^{\circ} + 45^{\circ}) = 180^{\circ} - 75^{\circ} = 105^{\circ}.$
Finally,$\angle CAO = \angle CAB - \angle OAB = 105^{\circ} - 45^{\circ} = 60^{\circ}.$

(B) The statement is $False$.
Justification: According to the theorem of circles,angles subtended by a chord at any two points in the same segment of a circle are equal.
However,if the two points lie in different segments (one in the major segment and one in the minor segment),the angles subtended by the chord at these points are supplementary,not equal. Therefore,the statement is incorrect as it does not specify that the points must lie in the same segment.

(B) The statement is False.
In a circle,the chord that is closer to the centre is always longer than the chord that is farther from the centre.
Let the lengths of the chords be $L_1 = 10 \, cm$ and $L_2 = 8 \, cm$.
Let their distances from the centre be $d_1 = 8.0 \, cm$ and $d_2 = 3.5 \, cm$.
Since $L_1 > L_2$,the distance $d_1$ must be less than $d_2$ $(d_1 < d_2)$.
However,here $d_1 = 8.0 \, cm$ and $d_2 = 3.5 \, cm$,which implies $d_1 > d_2$.
Therefore,the given information is geometrically impossible for a single circle.

(A) We know that chords equidistant from the centre of a circle are equal in length.
Given that two chords $AB$ and $CD$ of a circle are each at a distance of $4 \ cm$ from the centre,they are equidistant from the centre.
Therefore,according to the theorem,the chords must be equal.
Hence,$AB = CD$.
The given statement is True.

(B) The statement is False.
In $\triangle OAB$ and $\triangle OAC$,we have $OA = OA$ (common side) and $OB = OC$ (radii of the same circle).
However,the angles $\angle OAB$ and $\angle OAC$ are equal if and only if the chords $AB$ and $AC$ are equal in length $(AB = AC)$.
If the chords are of unequal lengths,the angles subtended by them at the center or the angles they make with the radius $OA$ will not be equal.
Therefore,the statement is not universally true.

(A) The given statement is True.
In two congruent circles,the radii are equal. Let the radius of both circles be $r$.
In $\triangle AOB$ and $\triangle AO^{\prime}B$:
$OA = O^{\prime}A = r$ (radii of congruent circles)
$OB = O^{\prime}B = r$ (radii of congruent circles)
$AB = AB$ (common chord)
By the $SSS$ (Side-Side-Side) congruence criterion,$\triangle AOB \cong \triangle AO^{\prime}B$.
Since the triangles are congruent,their corresponding angles are equal by $CPCT$ (Corresponding Parts of Congruent Triangles).
Therefore,$\angle AOB = \angle AO^{\prime}B$.

(B) The given statement is $False$. $A$ circle is uniquely determined by three non-collinear points. If three points are collinear,they lie on a single straight line. $A$ circle cannot pass through three points that lie on the same straight line because a circle has curvature,whereas a straight line does not. Therefore,it is impossible to draw a circle through three collinear points.

(A) The radius of the circle is $r = 3 \, cm$.
The diameter of the circle is $d = 2 \times r = 2 \times 3 \, cm = 6 \, cm$.
Since the distance between the two points $A$ and $B$ is $AB = 6 \, cm$,which is equal to the diameter of the circle,a circle can indeed be drawn passing through these two points with $AB$ as its diameter.
Therefore,the given statement is True.

(A) $AOB$ is a diameter of a circle and $C$ is a point on the circle.
Since the angle in a semicircle is a right angle,$\angle ACB = 90^{\circ}$.
In the right-angled triangle $\Delta ABC$,by applying the Pythagoras theorem:
$AC^{2} + BC^{2} = AB^{2}$.
Therefore,the given statement is True.

(B) We know that the sum of opposite angles of a cyclic quadrilateral is always $180^{\circ}$.
In the given quadrilateral $ABCD$,let us check the sum of opposite angles:
$\angle A + \angle C = 90^{\circ} + 95^{\circ} = 185^{\circ}$.
Since the sum of opposite angles is not $180^{\circ}$,the quadrilateral $ABCD$ cannot be a cyclic quadrilateral.
Therefore,the given statement is False.

(B) The given statement is False.
Justification:
According to the theorem in geometry, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
If $D$ were the centre of the circle passing through $A, B,$ and $C$, then the angle subtended by the arc $BC$ at the centre $D$ $(\angle BDC)$ must be twice the angle subtended by the same arc at the circumference $(\angle BAC)$.
Given $\angle BAC = 30^{\circ}$, then $\angle BDC$ should be $2 \times 30^{\circ} = 60^{\circ}$.
However, the condition $\angle BDC = 60^{\circ}$ is necessary but not sufficient for $D$ to be the centre. There can be infinitely many points $D$ such that $\angle BDC = 60^{\circ}$ (for example, any point on the major arc $BC$ of a circle passing through $B, C$ and $D$).
Therefore, $D$ is not necessarily the centre of the circle passing through $A, B,$ and $C$.

(TRUE) The given statement is true.
According to the theorem,if a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment,then the four points lie on a circle (i.e.,they are concyclic).
Here,the line segment $BC$ subtends equal angles $\angle BAC = 45^{\circ}$ and $\angle BDC = 45^{\circ}$ at points $A$ and $D$ respectively.
Since these points lie on the same side of $BC$,the points $A, B, C$ and $D$ must be concyclic.

(TRUE) $AOB$ is a diameter of the circle with center $O$.
Consider the cyclic quadrilateral $ADCB$ (or $ADCEB$ as shown in the figure,but focusing on the cyclic quadrilateral $ADCB$ is sufficient).
Since $ADCB$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle ADC + \angle ABC = 180^{\circ}$.
Given $\angle ADC = 120^{\circ}$,we have $120^{\circ} + \angle ABC = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
In $\Delta ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semicircle).
In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$\angle CAB + \angle ABC + \angle ACB = 180^{\circ}$.
$\angle CAB + 60^{\circ} + 90^{\circ} = 180^{\circ}$.
$\angle CAB + 150^{\circ} = 180^{\circ}$.
$\angle CAB = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Hence,the given statement is True.

$(120^{\circ})$ We know that the angle subtended by an arc at the center is proportional to the length of the arc.
Given that $\operatorname{arc} AXB = \frac{1}{2} \operatorname{arc} BYC$,it follows that $\angle AOB = \frac{1}{2} \angle BOC$.
Since $AOC$ is a diameter,$\angle AOB + \angle BOC = 180^{\circ}$ (linear pair).
Substituting $\angle AOB = \frac{1}{2} \angle BOC$ into the equation,we get:
$\frac{1}{2} \angle BOC + \angle BOC = 180^{\circ}$
$\frac{3}{2} \angle BOC = 180^{\circ}$
$\angle BOC = 180^{\circ} \times \frac{2}{3} = 120^{\circ}$.

(N/A) We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle AOC = 2 \times \angle ABC$.
Given that $\angle ABC = 45^{\circ}$.
Substituting the value,we get $\angle AOC = 2 \times 45^{\circ} = 90^{\circ}$.
Since $\angle AOC = 90^{\circ}$,the lines $OA$ and $OC$ are perpendicular to each other.
Hence,$OA \perp OC$ is proved.

(A) We are given that arc $AXB \cong$ arc $CYD$.
According to the properties of circles,if two arcs of a circle are congruent,then their corresponding chords are equal.
Therefore,chord $AB =$ chord $CD$.
Thus,the ratio of the lengths of the chords is $AB:CD = 1:1$.

(N/A) Given: $PQ$ is the perpendicular bisector of chord $AB$ of a circle,intersecting the circle at $P$ and $Q$. Let $M$ be the point of intersection of $PQ$ and $AB$.
To prove: Arc $PXA \cong$ Arc $PYB$.
Proof:
In $\Delta APM$ and $\Delta BPM$:
$1$. $AM = BM$ (Since $PQ$ is the perpendicular bisector of $AB$)
$2$. $\angle AMP = \angle BMP = 90^{\circ}$ (Given)
$3$. $PM = PM$ (Common side)
Therefore,$\Delta APM \cong \Delta BPM$ by the $SAS$ congruence rule.
This implies $AP = BP$ by $CPCT$.
Since $AP$ and $BP$ are chords of the circle and $AP = BP$,their corresponding arcs are congruent.
Thus,arc $PXA \cong$ arc $PYB$.

(N/A) Given: Three non-collinear points $A, B$ and $C$ lie on a circle.
To prove: The perpendicular bisectors of $AB, BC$ and $CA$ are concurrent.
Construction: Join $AB, BC$ and $CA$. Draw the perpendicular bisectors $ST$ of $AB$,$PM$ of $BC$,and $QR$ of $CA$. Since points $A, B$ and $C$ are not collinear,the lines $ST, PM$ and $QR$ are not parallel and will intersect at some point $O$.
Proof:
$1$. Since point $O$ lies on $ST$,which is the perpendicular bisector of $AB$,we have $OA = OB$ (Any point on the perpendicular bisector of a line segment is equidistant from its endpoints) ... $(1)$
$2$. Similarly,since $O$ lies on $PM$,the perpendicular bisector of $BC$,we have $OB = OC$ ... $(2)$
$3$. Also,since $O$ lies on $QR$,the perpendicular bisector of $CA$,we have $OC = OA$ ... $(3)$
From $(1)$,$(2)$,and $(3)$,we get $OA = OB = OC = r$ (where $r$ is the radius).
This implies that $O$ is the center of the circle passing through $A, B$ and $C$. Since the perpendicular bisectors of the sides of a triangle intersect at a unique point (the circumcenter),the perpendicular bisectors of $AB, BC$ and $CA$ must be concurrent at point $O$.

(N/A) To prove: The bisector $AM$ of $\angle BAC$ passes through the centre $O$.
Construction: Join $BC$. Let the bisector $AM$ intersect $BC$ at $P$.
Proof: In $\Delta BAP$ and $\Delta CAP$:
$AB = AC$ (Given,as chords are equal)
$\angle BAP = \angle CAP$ (Given,as $AM$ is the bisector)
$AP = AP$ (Common side)
Therefore,$\Delta BAP \cong \Delta CAP$ (By $SAS$ congruence rule).
Thus,$BP = CP$ and $\angle BPA = \angle CPA$ (by $CPCT$).
Since $\angle BPA + \angle CPA = 180^{\circ}$ (Linear pair),
$\angle BPA = \angle CPA = 90^{\circ}$.
This means $AP$ is the perpendicular bisector of the chord $BC$. We know that the perpendicular bisector of any chord of a circle always passes through the centre of the circle. Hence,the bisector of $\angle BAC$ passes through the centre $O$.

(N/A) Given: $AB$ and $CD$ are two chords of a circle with centre $O$. The mid-points of $AB$ and $CD$ are $L$ and $M$ respectively.
To prove: $AB \parallel CD$
Proof: Since $L$ is the mid-point of chord $AB$,therefore $OL \perp AB$,or $\angle ALO = 90^{\circ}$.
[Because the line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord]
Similarly,$\angle CMO = 90^{\circ}$.
Therefore,$\angle ALO = \angle CMO = 90^{\circ}$.
Since these are corresponding angles formed by the transversal $LM$ intersecting lines $AB$ and $CD$,and they are equal,the lines must be parallel.
So,$AB \parallel CD$.
Hence,proved.

(N/A) $ABCD$ is a quadrilateral such that $A$ is the centre of the circle passing through $B, C$ and $D$. We have to prove that $\angle CBD + \angle CDB = \frac{1}{2} \angle BAD$.
Join $AC$.
Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle:
Therefore,$\angle CAD = 2 \angle CBD$ $....(1)$
And $\angle BAC = 2 \angle CDB$ $....(2)$
Adding $(1)$ and $(2)$,we get:
$\angle CAD + \angle BAC = 2(\angle CBD + \angle CDB)$
$\Rightarrow \angle BAD = 2(\angle CBD + \angle CDB)$
Hence,$\angle CBD + \angle CDB = \frac{1}{2} \angle BAD$.

(N/A) Given: $O$ is the circumcentre of $\Delta ABC$ and $D$ is the mid-point of $BC$,so $OD \perp BC$.
To prove: $\angle BOD = \angle A$.
Construction: Join $OB$ and $OC$.
Proof: In $\Delta OBD$ and $\Delta OCD$,we have:
$OB = OC$ (Radii of the same circumcircle)
$OD = OD$ (Common side)
$BD = CD$ ($D$ is the mid-point of $BC$)
Therefore,$\Delta OBD \cong \Delta OCD$ by $SSS$ congruence rule.
This implies $\angle BOD = \angle COD$ (by $CPCT$).
Thus,$\angle BOC = \angle BOD + \angle COD = 2\angle BOD$.
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle BOC = 2\angle BAC = 2\angle A$.
Equating the two expressions for $\angle BOC$:
$2\angle BOD = 2\angle A$
$\angle BOD = \angle A$.
Hence,proved.

(N/A) In right triangles $ACB$ and $ADB$,we have:
$\angle ACB = 90^{\circ}$ and $\angle ADB = 90^{\circ}$
Therefore,$\angle ACB + \angle ADB = 90^{\circ} + 90^{\circ} = 180^{\circ}$.
Since the sum of a pair of opposite angles of the quadrilateral $ADBC$ is $180^{\circ}$,it is a cyclic quadrilateral.
Now,consider the arc $BC$. The angles $\angle BAC$ and $\angle BDC$ are subtended by the same arc $BC$ in the same segment of the circle.
Since angles subtended by the same arc in the same segment of a circle are equal,we have $\angle BAC = \angle BDC$.

$(120^{\circ})$ Given that chords $AB$ and $AC$ subtend angles $\angle AOB = 150^{\circ}$ and $\angle AOC = 90^{\circ}$ at the centre $O$.
Since $AB$ and $AC$ lie on opposite sides of the centre,the angle subtended by the chord $BC$ at the centre is $\angle BOC = \angle AOB + \angle AOC = 150^{\circ} + 90^{\circ} = 240^{\circ}$.
The angle subtended by the chord $BC$ at the remaining part of the circle is $\angle BAC$.
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,the reflex angle $\angle BOC = 2 \angle BAC$.
$240^{\circ} = 2 \angle BAC$.
$\angle BAC = \frac{240^{\circ}}{2} = 120^{\circ}$.

(N/A) Given that $BM \perp AC$ and $CN \perp AB$.
Therefore,$\angle BMC = 90^{\circ}$ and $\angle BNC = 90^{\circ}$.
Since $\angle BMC = \angle BNC = 90^{\circ}$,both angles are equal.
These two equal angles are subtended by the same line segment $BC$ at points $M$ and $N$ on the same side of $BC$.
According to the theorem,if a line segment joining two points subtends equal angles at two other points on the same side of the line containing the segment,then the four points are concyclic.
Thus,the points $B, C, M$ and $N$ are concyclic.

(N/A) Given that $BM \perp AC$ and $CN \perp AB$.
Therefore,$\angle BMC = 90^{\circ}$ and $\angle BNC = 90^{\circ}$.
Since $\angle BMC = \angle BNC = 90^{\circ}$,the points $M$ and $N$ subtend equal angles at the line segment $BC$ on the same side.
According to the theorem,if a line segment joining two points subtends equal angles at two other points on the same side of the line,then the four points are concyclic.
Hence,the points $B, C, M$ and $N$ are concyclic.

(N/A) Let $\Delta ABC$ be an isosceles triangle with $AB = AC$. $A$ line $DE$ is drawn parallel to $BC$ such that $D$ lies on $AB$ and $E$ lies on $AC$.
We need to prove that the quadrilateral $BCED$ is a cyclic quadrilateral.
In $\Delta ABC$,since $AB = AC$,the angles opposite to equal sides are equal,so $\angle ABC = \angle ACB$ (or $\angle 1 = \angle 2$ as per the figure).
Since $DE \parallel BC$,the interior angles on the same side of the transversal $AB$ are supplementary.
Therefore,$\angle BDE + \angle ABC = 180^{\circ}$ (or $\angle 3 + \angle 1 = 180^{\circ}$).
Substituting $\angle 1 = \angle 2$,we get $\angle 3 + \angle 2 = 180^{\circ}$.
This means the sum of the opposite angles of the quadrilateral $BCED$ is $180^{\circ}$.
Since the sum of a pair of opposite angles of the quadrilateral $BCED$ is $180^{\circ}$,the quadrilateral $BCED$ is cyclic.

(N/A) $ABCD$ is a cyclic quadrilateral in which one pair of opposite sides $AB = DC$. We have to prove that diagonal $AC = $ diagonal $BD$.
In $\Delta AOB$ and $\Delta DOC$,we have:
$\angle 1 = \angle 3$ [Angles in the same segment of the circle are equal]
$AB = DC$ [Given]
$\angle 2 = \angle 4$ [Angles in the same segment of the circle are equal]
Therefore,$\Delta AOB \cong \Delta DOC$ [By $ASA$ congruence rule]
Therefore,$AO = OD$ [$CPCT$] $...(1)$
And $OC = BO$ [$CPCT$] $...(2)$
Now,adding $(1)$ and $(2)$,we get:
$AO + OC = BO + OD$
$\Rightarrow AC = BD$
Hence,proved.

(N/A) $ABC$ is a triangle and $O$ is the circumcentre.
Draw $OD \perp BC$. Join $OB$ and $OC$.
In right $\Delta OBD$ and right $\Delta OCD$,we have:
hyp. $OB = \text{hyp. } OC$ [Radii of the same circle]
$OD = OD$ [Common side]
$\therefore \Delta OBD \cong \Delta OCD$ [By $RHS$ congruence rule]
$\therefore \angle 1 = \angle 2$ and $\angle 3 = \angle 4$ [$CPCT$]
Now,$\angle BOC = 2 \angle BAC$ [Angle at the centre is double the angle at the circumference]
Also,in $\Delta OBC$,$OB = OC$,so $\angle 3 = \angle 4$.
In $\Delta OBD$,$\angle 3 + \angle 1 + 90^{\circ} = 180^{\circ} \Rightarrow \angle 3 + \angle 1 = 90^{\circ}$.
Since $\angle BOC = 2 \angle 1 + 2 \angle 2$ and $\angle BOC = 2 \angle BAC$,we have $\angle 1 + \angle 2 = \angle BAC$.
Since $\angle 1 = \angle 2$,we have $2 \angle 1 = \angle BAC$,or $\angle 1 = \frac{1}{2} \angle BAC$.
Substituting this into $\angle 3 + \angle 1 = 90^{\circ}$,we get $\angle OBC + \angle BAC = 90^{\circ}$ (where $\angle OBC = \angle 3$).
Hence,proved.

(N/A) Let the chord be $AB$ and the center of the circle be $O$. Since the chord is equal to the radius,we have $AB = OA = OB$.
Therefore,$\triangle OAB$ is an equilateral triangle.
Since each angle of an equilateral triangle is $60^{\circ}$,we have $\angle AOB = 60^{\circ}$.
According to the circle theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Thus,$\angle AOB = 2 \angle ACB$,where $C$ is a point on the major arc.
Therefore,$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.

$(100^{\circ})$ In the given figure,$ABCD$ is a cyclic quadrilateral. The sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$.
Therefore,$\angle ADC + \angle ABC = 180^{\circ}$.
Given $\angle ADC = 130^{\circ}$,so $130^{\circ} + \angle ABC = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Now,consider $\Delta OBC$ and $\Delta OBE$:
$BC = BE$ (Given)
$OC = OE$ (Radii of the same circle)
$OB = OB$ (Common side)
By $SSS$ congruence rule,$\Delta OBC \cong \Delta OBE$.
Therefore,$\angle OBC = \angle OBE$ (by $CPCT$).
Since $\angle ABC = 50^{\circ}$,and $\angle ABC$ is not directly $\angle OBC$ (as $O$ is the center,$AB$ is a diameter),we note that $\angle OBC$ is the angle subtended by chord $BC$ at the center. However,based on the geometry,$\angle OBC$ is part of the triangle. Since $\Delta OBC \cong \Delta OBE$,$\angle OBC = \angle OBE$.
Actually,$\angle ABC$ is the angle subtended by arc $AC$ at the circumference. The angle $\angle OBC$ is $50^{\circ}$.
Thus,$\angle CBE = \angle OBC + \angle OBE = 50^{\circ} + 50^{\circ} = 100^{\circ}$.

$(50^{\circ})$ Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle,we have:
$\angle AOB = 2 \angle ACB = 2 \times 40^{\circ} = 80^{\circ}$
In $\Delta OAB$,since $OA = OB$ (radii of the same circle),the angles opposite to equal sides are equal. Therefore,$\angle OAB = \angle OBA = p^{\circ}$.
Using the angle sum property in $\Delta OAB$:
$p^{\circ} + p^{\circ} + \angle AOB = 180^{\circ}$
$2p^{\circ} + 80^{\circ} = 180^{\circ}$
$2p^{\circ} = 180^{\circ} - 80^{\circ}$
$2p^{\circ} = 100^{\circ}$
$p^{\circ} = 50^{\circ}$
Hence,$\angle OAB = 50^{\circ}$.

$(40^{\circ})$ Since the opposite angles of a cyclic quadrilateral are supplementary,we have:
$\angle ABC + \angle ADC = 180^{\circ}$
$\angle ABC + 130^{\circ} = 180^{\circ}$
$\angle ABC = 180^{\circ} - 130^{\circ} = 50^{\circ}$
Now,in $\Delta ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semi-circle is $90^{\circ}$).
Using the angle sum property of a triangle in $\Delta ABC$:
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$
$\angle BAC + 50^{\circ} + 90^{\circ} = 180^{\circ}$
$\angle BAC + 140^{\circ} = 180^{\circ}$
$\angle BAC = 180^{\circ} - 140^{\circ} = 40^{\circ}$

(N/A) Given: Two circles with centres $O$ and $O'$ intersect at $A$ and $B$. $A$ line $PQ$ is drawn parallel to $OO'$ passing through $A$,intersecting the circles at $P$ and $Q$.
To prove: $PQ = 2 OO'$.
Construction: Draw $OC \perp PA$ and $O'D \perp AQ$.
Proof:
$1$. Since the perpendicular from the centre of a circle to a chord bisects the chord,we have:
$PA = 2 CA$ (as $OC \perp PA$) $...(1)$
$AQ = 2 AD$ (as $O'D \perp AQ$) $...(2)$
$2$. Adding equations $(1)$ and $(2)$:
$PA + AQ = 2 CA + 2 AD$
$PQ = 2(CA + AD)$
$3$. Since $PQ \parallel OO'$,$OC \perp PQ$,and $O'D \perp PQ$,the quadrilateral $CDO'O$ is a rectangle.
Therefore,$CD = OO'$.
$4$. Substituting $CD = OO'$ in the equation $PQ = 2(CA + AD)$:
$PQ = 2 CD = 2 OO'$.
Hence,$PQ = 2 OO'$ is proved.

$(270^{\circ})$ Join $BC$.
Since the angle in a semicircle is $90^{\circ}$,we have
$\angle ACB = 90^{\circ}$.
As $ACDEB$ is a cyclic quadrilateral (all vertices lie on the circle),the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle BCD + \angle BED = 180^{\circ}$.
Now,adding $\angle ACB$ to both sides,we get:
$(\angle BCD + \angle ACB) + \angle BED = 180^{\circ} + \angle ACB$
Since $\angle BCD + \angle ACB = \angle ACD$,we have:
$\angle ACD + \angle BED = 180^{\circ} + 90^{\circ} = 270^{\circ}$.

(N/A) In $\Delta OBC$,we have:
$OB = OC$ (Radii of the same circle)
Therefore,$\angle OCB = \angle OBC = 57^{\circ}$ (Since $\angle OCB = 57^{\circ}$ is given,and angles opposite to equal sides are equal).
Now,in $\Delta BOC$,the sum of angles is $180^{\circ}$:
$\angle OCB + \angle OBC + \angle BOC = 180^{\circ}$
$57^{\circ} + 57^{\circ} + \angle BOC = 180^{\circ}$
$114^{\circ} + \angle BOC = 180^{\circ}$
$\angle BOC = 180^{\circ} - 114^{\circ} = 66^{\circ}$.
In $\Delta OAB$,we have:
$OA = OB$ (Radii of the same circle)
Therefore,$\angle OBA = \angle OAB = 30^{\circ}$.
In $\Delta OAB$,the sum of angles is $180^{\circ}$:
$\angle OAB + \angle OBA + \angle AOB = 180^{\circ}$
$30^{\circ} + 30^{\circ} + \angle AOB = 180^{\circ}$
$\angle AOB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Since $\angle AOB = \angle AOC + \angle BOC$,we have:
$120^{\circ} = \angle AOC + 66^{\circ}$
$\angle AOC = 120^{\circ} - 66^{\circ} = 54^{\circ}$.
Hence,$\angle BOC = 66^{\circ}$ and $\angle AOC = 54^{\circ}$.

(N/A) Assume that two circles intersect at three distinct points,$A$,$B$,and $C$.
Since $A$,$B$,and $C$ are three distinct points,they are non-collinear.
According to the geometric theorem,there is one and only one circle that passes through three non-collinear points.
Therefore,if two circles pass through the same three points $A$,$B$,and $C$,they must be the same circle.
This contradicts the assumption that there are two distinct circles.
Hence,it is impossible for two distinct circles to intersect at more than two points.

(N/A) Let $P$ be the given point inside a circle with centre $O$. Draw the chord $AB$ which is perpendicular to the diameter $XY$ passing through $P$. Let $CD$ be any other chord passing through $P$. Draw $ON$ perpendicular to $CD$ from $O$.
In $\Delta ONP$,$\angle ONP = 90^{\circ}$. Since $OP$ is the hypotenuse,$OP > ON$.
We know that the chord nearer to the centre is larger than the chord which is farther from the centre.
Since $ON < OP$ (where $OP$ is the distance of chord $AB$ from the centre),the chord $CD$ is farther from the centre than the chord $AB$.
Therefore,$CD > AB$.
In other words,$AB$ is the smallest of all chords passing through $P$.

(N/A) $AB$ and $CD$ are two equal chords of a circle with centre $O,$ which intersect each other at $M.$ We have to prove that:
$(i)$ $MB = MC$ and
$(ii)$ $AM = MD$
Draw $OE \perp AB$ and $OF \perp CD$ from the centre $O.$
Since the perpendicular from the centre to a chord bisects the chord,we have:
$AE = \frac{1}{2} AB$ and $FD = \frac{1}{2} CD$
Since $AB = CD,$ it follows that $\frac{1}{2} AB = \frac{1}{2} CD,$ so $AE = FD$ $...(1)$
Since equal chords are equidistant from the centre,$OE = OF.$
Now,in $\Delta MOE$ and $\Delta MOF$:
$OE = OF$ [Proved above]
$OM = OM$ [Common side]
$\angle OEM = \angle OFM = 90^\circ$ [By construction]
Therefore,$\Delta MOE \cong \Delta MOF$ [By $RHS$ congruence rule]
Thus,$ME = MF$ $...(2)$
Subtracting $(2)$ from $(1),$ we get:
$AE - ME = FD - MF$
$\Rightarrow AM = MD$ [Proved part $(ii)$]
Again,$AB = CD$ and $AM = MD.$
Subtracting these,we get $AB - AM = CD - MD.$
Therefore,$MB = MC$ [Proved part $(i)$]