Cyclic Quadrilaterals Questions in English

(A) We know that angles in the same segment of a circle are equal.
Therefore,$\angle CAD = \angle DBC = 55^{\circ}$.
Now,the total angle $\angle DAB = \angle CAD + \angle BAC$.
Substituting the values,$\angle DAB = 55^{\circ} + 45^{\circ} = 100^{\circ}$.
Since $ABCD$ is a cyclic quadrilateral,the sum of its opposite angles is $180^{\circ}$.
Thus,$\angle DAB + \angle BCD = 180^{\circ}$.
Substituting the value of $\angle DAB$,we get $100^{\circ} + \angle BCD = 180^{\circ}$.
Therefore,$\angle BCD = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

(N/A) In the figure,$ABCD$ is a quadrilateral in which the angle bisectors of internal angles $A, B, C$ and $D$ intersect to form a quadrilateral $EFGH$.
Now,in $\triangle AEB$,$\angle FEH = \angle AEB = 180^{\circ} - (\angle EAB + \angle EBA)$
$= 180^{\circ} - \frac{1}{2}(\angle A + \angle B)$
Similarly,in $\triangle CGD$,$\angle FGH = \angle CGD = 180^{\circ} - (\angle GCD + \angle GDC)$
$= 180^{\circ} - \frac{1}{2}(\angle C + \angle D)$
Adding these two equations:
$\angle FEH + \angle FGH = 180^{\circ} - \frac{1}{2}(\angle A + \angle B) + 180^{\circ} - \frac{1}{2}(\angle C + \angle D)$
$= 360^{\circ} - \frac{1}{2}(\angle A + \angle B + \angle C + \angle D)$
Since the sum of angles in a quadrilateral is $360^{\circ}$,we have:
$= 360^{\circ} - \frac{1}{2}(360^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}$
Since the sum of opposite angles of quadrilateral $EFGH$ is $180^{\circ}$,it is a cyclic quadrilateral.

(A) Since angles in the same segment of a circle are equal,we have $\angle BAC = \angle BDC$.
Given $\angle BAC = 30^{\circ}$,therefore $\angle BDC = 30^{\circ}$.
In $\Delta BCD$,the sum of angles is $180^{\circ}$:
$\angle BCD + \angle DBC + \angle BDC = 180^{\circ}$
$\angle BCD + 70^{\circ} + 30^{\circ} = 180^{\circ}$
$\angle BCD = 180^{\circ} - 100^{\circ} = 80^{\circ}$.
Now,in $\Delta ABC$,since $AB = BC$,the angles opposite to equal sides are equal,so $\angle BCA = \angle BAC = 30^{\circ}$.
Since $\angle BCD = \angle BCA + \angle ECD$,we have:
$80^{\circ} = 30^{\circ} + \angle ECD$
$\angle ECD = 80^{\circ} - 30^{\circ} = 50^{\circ}$.

(N/A) Given: $ABCD$ is a cyclic quadrilateral where diagonals $AC$ and $BD$ are diameters of the circle.
Since $AC$ and $BD$ are diameters,they pass through the center $O$ and are equal in length $(AC = BD)$.
We know that an angle in a semicircle is a right angle.
Since $AC$ is a diameter,$\angle ABC = 90^{\circ}$ and $\angle ADC = 90^{\circ}$.
Since $BD$ is a diameter,$\angle BAD = 90^{\circ}$ and $\angle BCD = 90^{\circ}$.
Thus,all interior angles of the quadrilateral $ABCD$ are $90^{\circ}$.
$A$ quadrilateral with all angles equal to $90^{\circ}$ is a rectangle.
Therefore,$ABCD$ is a rectangle.

(N/A) We have a trapezium $ABCD$ such that $AB || CD$ and $AD = BC$.
Let us draw $BE || AD$ such that $ABED$ is a parallelogram.
Since the opposite angles of a parallelogram are equal,
$\angle BAD = \angle BED$ ..... $(1)$
and $AD = BE$ [Opposite sides of a parallelogram] ...... $(2)$
But $AD = BC$ [Given] ......... $(3)$
From $(2)$ and $(3)$,we have
$BE = BC$
$\Rightarrow \angle BEC = \angle BCE$ [Angles opposite to equal sides are equal]
Now,$\angle BED + \angle BEC = 180^{\circ}$ [Linear pair]
Since $\angle BED = \angle BAD$ and $\angle BEC = \angle BCE$,we have
$\angle BAD + \angle BCE = 180^{\circ}$
In quadrilateral $ABCD$,$\angle BAD + \angle BCD = \angle BAD + (\angle BCE + \angle ECD)$. Since $ABED$ is a parallelogram,$\angle ECD = \angle BAD$. Thus,the sum of opposite angles is $180^{\circ}$.
Therefore,$ABCD$ is a cyclic quadrilateral.

(N/A) Let $ABCD$ be a cyclic parallelogram.
Since $ABCD$ is a cyclic quadrilateral,the sum of its opposite angles is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$ .... $(1)$
In a parallelogram,opposite angles are equal,so $\angle A = \angle C$ .... $(2)$
Substituting $(2)$ in $(1)$,we get:
$\angle A + \angle A = 180^{\circ}$
$2\angle A = 180^{\circ}$
$\angle A = 90^{\circ}$
Since $\angle A = \angle C$,we have $\angle C = 90^{\circ}$.
Similarly,for the other pair of opposite angles,$\angle B + \angle D = 180^{\circ}$ and $\angle B = \angle D$,which implies $\angle B = \angle D = 90^{\circ}$.
Since all angles of the parallelogram $ABCD$ are $90^{\circ}$,$ABCD$ is a rectangle.

(N/A) Given: $ABCD$ is a parallelogram. $A$ circle passes through $A, B$ and $C$,intersecting $CD$ at $E$.
Step $1$: Since $ABCE$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle AEC + \angle B = 180^{\circ}$ --- $(1)$
Step $2$: In parallelogram $ABCD$,opposite angles are equal.
Therefore,$\angle D = \angle B$ --- $(2)$
Step $3$: From equations $(1)$ and $(2)$,we get:
$\angle AEC + \angle D = 180^{\circ}$ --- $(3)$
Step $4$: Since $D, E, C$ lie on a straight line,$\angle AEC$ and $\angle AED$ form a linear pair.
Therefore,$\angle AEC + \angle AED = 180^{\circ}$ --- $(4)$
Step $5$: Comparing equations $(3)$ and $(4)$:
$\angle D = \angle AED$
Step $6$: In $\Delta ADE$,since the base angles $\angle D$ and $\angle AED$ are equal,the sides opposite to these angles must be equal.
Therefore,$AE = AD$. Hence proved.

(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 $180^{\circ}$:
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$
$\angle BAC + 40^{\circ} + 90^{\circ} = 180^{\circ}$
$\angle BAC + 130^{\circ} = 180^{\circ}$
$\angle BAC = 180^{\circ} - 130^{\circ} = 50^{\circ}$

(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.

(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 $\angle BAC = \angle BDC$.

(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.

$(100^{\circ})$ 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,$\angle CBE = \angle OBC + \angle OBE = 50^{\circ} + 50^{\circ} = 100^{\circ}$.

$(40^{\circ})$ 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: $ABCD$ is a trapezium in which $AD \parallel BC$ and its non-parallel sides $AB$ and $DC$ are equal,i.e.,$AB = DC$.
To prove: Trapezium $ABCD$ is cyclic.
Construction: Draw $AM \perp BC$ and $DN \perp BC$.
Proof: In right-angled triangles $\Delta AMB$ and $\Delta DNC$:
$\angle AMB = \angle DNC = 90^{\circ}$ (By construction)
$AB = DC$ (Given)
$AM = DN$ (Distance between two parallel lines is constant)
Therefore,$\Delta AMB \cong \Delta DNC$ (By $RHS$ congruence rule).
Thus,$\angle B = \angle C$ $(CPCT)$.
Also,$\angle BAM = \angle CDN$ $(CPCT)$.
Since $\angle AMB = 90^{\circ}$ and $\angle DNC = 90^{\circ}$,we have $\angle MAB = 90^{\circ} - \angle B$ and $\angle NDC = 90^{\circ} - \angle C$. Since $\angle B = \angle C$,then $\angle MAB = \angle NDC$.
Now,$\angle BAD = \angle BAM + \angle MAD = \angle BAM + 90^{\circ}$ and $\angle CDA = \angle CDN + \angle NDA = \angle CDN + 90^{\circ}$.
Since $\angle BAM = \angle CDN$,it follows that $\angle BAD = \angle CDA$.
In quadrilateral $ABCD$,$\angle B + \angle C + \angle CDA + \angle BAD = 360^{\circ}$.
Substituting $\angle C = \angle B$ and $\angle CDA = \angle BAD$,we get $2\angle B + 2\angle BAD = 360^{\circ}$,which implies $\angle B + \angle BAD = 180^{\circ}$.
Since the sum of a pair of opposite angles is $180^{\circ}$,the trapezium $ABCD$ is cyclic.

(N/A) $ABCD$ is a parallelogram. $A$ circle passes through $A$ and $B$,intersecting $AD$ at $P$ and $BC$ at $Q$. We need to prove that $P, Q, C,$ and $D$ are concyclic.
$1$. Since $ABQP$ is a cyclic quadrilateral,the exterior angle formed by extending side $AP$ to $D$ is equal to the interior opposite angle.
Let $\angle QPD$ be the exterior angle at $P$. Then,$\angle QPD = \angle ABQ$.
$2$. In parallelogram $ABCD$,$AD \parallel BC$. Since $AB$ is a transversal,$\angle A + \angle B = 180^{\circ}$. However,considering the parallel lines $AB$ and $DC$ with transversal $BC$,we have $\angle B + \angle C = 180^{\circ}$ (consecutive interior angles).
$3$. Since $\angle QPD = \angle ABQ$ (exterior angle property of cyclic quadrilateral $ABQP$),and we know $\angle ABQ + \angle C = 180^{\circ}$,we can substitute $\angle QPD$ for $\angle ABQ$.
$4$. Therefore,$\angle QPD + \angle C = 180^{\circ}$.
$5$. Since the sum of the opposite angles of quadrilateral $PDCQ$ is $180^{\circ}$,$PDCQ$ is a cyclic quadrilateral.
Hence,the points $P, Q, C,$ and $D$ are concyclic.

(N/A) Let the bisectors of opposite angles $\angle A$ and $\angle C$ of a cyclic quadrilateral $ABCD$ intersect the circle at points $P$ and $Q$ respectively.
We need to prove that $PQ$ is a diameter of the circle.
Join $AQ$ and $DQ$.
Since the opposite angles of a cyclic quadrilateral are supplementary,in cyclic quadrilateral $ABCD$,we have:
$\angle DAB + \angle DCB = 180^{\circ}$
Dividing by $2$,we get:
$\frac{1}{2} \angle DAB + \frac{1}{2} \angle DCB = 90^{\circ}$
Let $\angle 1 = \frac{1}{2} \angle DAB$ and $\angle 2 = \frac{1}{2} \angle DCB$. Thus,$\angle 1 + \angle 2 = 90^{\circ}$.
Since $\angle 2$ and $\angle 3$ are angles in the same segment subtended by the chord $QD$,we have $\angle 2 = \angle 3$.
Substituting this into the equation,we get $\angle 1 + \angle 3 = 90^{\circ}$.
This implies $\angle PAQ = 90^{\circ}$.
Since the angle subtended by $PQ$ at the circumference is $90^{\circ}$,$PQ$ must be a diameter of the circle.

(B) We know that angles in the same segment of a circle are equal.
Therefore,$\angle BDC = \angle BAC$.
Given $\angle BAC = 50^{\circ}$,so $\angle BDC = 50^{\circ}$.
Now,$\angle ADC = \angle ADB + \angle BDC$.
Substituting the values,$\angle ADC = 45^{\circ} + 50^{\circ} = 95^{\circ}$.
Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle ABC + \angle ADC = 180^{\circ}$.
$\angle ABC + 95^{\circ} = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 95^{\circ} = 85^{\circ}$.

(N/A) Given: $ABCD$ is a cyclic quadrilateral with $AB \parallel CD$.
To prove: $BC = AD$.
Proof:
$1$. Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$\angle A + \angle C = 180^{\circ}$ and $\angle B + \angle D = 180^{\circ}$.
$2$. Since $AB \parallel CD$,the sum of consecutive interior angles is $180^{\circ}$. Thus,$\angle A + \angle D = 180^{\circ}$.
$3$. From $\angle A + \angle C = 180^{\circ}$ and $\angle A + \angle D = 180^{\circ}$,we get $\angle C = \angle D$.
$4$. Similarly,since $AB \parallel CD$,$\angle B + \angle C = 180^{\circ}$. Given $\angle B + \angle D = 180^{\circ}$,we get $\angle C = \angle D$.
$5$. In a cyclic quadrilateral,if one pair of opposite sides is parallel,it must be an isosceles trapezium. In an isosceles trapezium,the non-parallel sides are equal.
$6$. Therefore,$BC = AD$.

(N/A) $1$. In $\triangle ABC$,we are given $AB = AC$. Therefore,$\angle ABC = \angle ACB$ (angles opposite to equal sides are equal).
$2$. Since $B, C, Y, X$ are concyclic points,$BCYX$ is a cyclic quadrilateral.
$3$. In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$\angle AXY + \angle AC B = 180^{\circ}$ is not directly useful,but rather the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
$4$. Specifically,for cyclic quadrilateral $BCYX$,the exterior angle $\angle AXY = \angle ACB$.
$5$. Since $\angle AXY = \angle ACB$ and we know $\angle ABC = \angle ACB$,it follows that $\angle AXY = \angle ABC$.
$6$. These are corresponding angles formed by the transversal $AB$ intersecting lines $XY$ and $BC$.
$7$. Since the corresponding angles are equal,$XY \parallel BC$.

(D) $1$. In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$\angle A + \angle C = 180^{\circ}$ and $\angle B + \angle D = 180^{\circ}$.
$2$. Given $\angle B = 80^{\circ}$,we have $\angle D = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
$3$. Since $AB \parallel CD$,the sum of interior angles on the same side of the transversal is $180^{\circ}$. Thus,$\angle A + \angle D = 180^{\circ}$.
$4$. Substituting $\angle D = 100^{\circ}$,we get $\angle A + 100^{\circ} = 180^{\circ}$,which implies $\angle A = 80^{\circ}$.
$5$. Finally,using $\angle A + \angle C = 180^{\circ}$,we get $80^{\circ} + \angle C = 180^{\circ}$,so $\angle C = 100^{\circ}$.
$6$. Therefore,the angles are $\angle A = 80^{\circ}, \angle C = 100^{\circ}, \angle D = 100^{\circ}$.

(D) In a cyclic quadrilateral,angles subtended by the same arc at the circumference are equal.
Since $ABCD$ is a cyclic quadrilateral,the angles subtended by the same arc $AB$ at the circumference are $\angle ACB$ and $\angle ADB$.
Therefore,$\angle ACB = \angle ADB = 78^{\circ}$.
Now,consider the triangle $\triangle ABC$. The sum of the interior angles of a triangle is $180^{\circ}$.
Thus,$\angle ABC + \angle BAC + \angle ACB = 180^{\circ}$.
Substituting the given values: $\angle ABC + 52^{\circ} + 78^{\circ} = 180^{\circ}$.
$\angle ABC + 130^{\circ} = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 130^{\circ} = 50^{\circ}$.

(A) $1$. In a cyclic quadrilateral $ABCD$,the sum of opposite angles is $180^{\circ}$.
$2$. Therefore,$\angle ABC + \angle ADC = 180^{\circ}$.
$3$. Given $\angle ADC = 150^{\circ}$,we have $\angle ABC = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
$4$. Since $AB$ is the diameter,the angle subtended by the diameter at the circumference is $90^{\circ}$. Thus,$\angle ACB = 90^{\circ}$.
$5$. In $\triangle ABC$,the sum of angles is $180^{\circ}$.
$6$. $\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$.
$7$. $\angle BAC + 30^{\circ} + 90^{\circ} = 180^{\circ}$.
$8$. $\angle BAC = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

(B) In a cyclic quadrilateral $PQRS$,angles subtended by the same arc at the circumference are equal.
Since $\angle QPR$ and $\angle QSR$ subtend the same arc $QR$,we have $\angle QSR = \angle QPR = 40^{\circ}$.
Given $\angle PSQ = 70^{\circ}$,the total angle $\angle PSR = \angle PSQ + \angle QSR = 70^{\circ} + 40^{\circ} = 110^{\circ}$.
In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle PQR + \angle PSR = 180^{\circ}$.
$\angle PQR + 110^{\circ} = 180^{\circ}$.
$\angle PQR = 180^{\circ} - 110^{\circ} = 70^{\circ}$.

(C) $1$. In cyclic quadrilateral $ABDE$,the exterior angle is equal to the interior opposite angle. Thus,$\angle CDE = \angle BAE = 65^{\circ}$. Since $CDE$ is a straight line,$\angle c = 180^{\circ} - 65^{\circ} = 115^{\circ}$ is incorrect based on the diagram geometry. Let's re-evaluate: In $\triangle ACE$,$\angle C = 40^{\circ}$ and $\angle CAE = 65^{\circ}$. Thus,$\angle AEC = 180^{\circ} - (40^{\circ} + 65^{\circ}) = 75^{\circ}$.
$2$. Since $ABDE$ is a cyclic quadrilateral,the exterior angle $\angle CDE = \angle BAE = 65^{\circ}$. Therefore,$c = 180^{\circ} - 65^{circ} = 115^{\circ}$ is not correct as $c$ is an interior angle of the cyclic quadrilateral. Actually,$\angle BDE = 180^{\circ} - 65^{\circ} = 115^{\circ}$.
$3$. From the cyclic property,$\angle BDE + \angle BAE = 180^{\circ} \implies \angle BDE = 180^{\circ} - 65^{\circ} = 115^{\circ}$.
$4$. $\angle d = \angle AED = 180^{\circ} - 75^{\circ} = 105^{\circ}$ is wrong. Let's use $\triangle ACD$: $\angle CAD = 65^{\circ}, \angle ACD = 40^{\circ} \implies \angle ADC = 180^{\circ} - 105^{\circ} = 75^{\circ}$.
$5$. Since $ABDE$ is cyclic,$\angle BDE = 180^{\circ} - 65^{\circ} = 115^{\circ}$. Thus,$c = 180^{\circ} - 115^{\circ} = 65^{\circ}$.
$6$. $\angle a = 180^{\circ} - \angle BDE = 180^{\circ} - 30^{\circ} = 150^{\circ}$.
$7$. In $\triangle FDE$,$\angle FDE = 180^{\circ} - 115^{\circ} = 65^{\circ}$ and $\angle FED = 180^{\circ} - 105^{\circ} = 75^{\circ}$. Thus,$b = 180^{\circ} - (65^{\circ} + 105^{\circ}) = 10^{\circ}$.
$8$. Therefore,$a=150^{\circ}, b=10^{\circ}, c=65^{\circ}, d=75^{\circ}$.

(D) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Given $\angle A : \angle C = 4 : 5$,let $\angle A = 4x$ and $\angle C = 5x$.
Since $\angle A + \angle C = 180^{\circ}$,we have $4x + 5x = 180^{\circ} \Rightarrow 9x = 180^{\circ} \Rightarrow x = 20^{\circ}$.
Thus,$\angle A = 4(20^{\circ}) = 80^{\circ}$ and $\angle C = 5(20^{\circ}) = 100^{\circ}$.
Given $\angle B : \angle D = 5 : 7$,let $\angle B = 5y$ and $\angle D = 7y$.
Since $\angle B + \angle D = 180^{\circ}$,we have $5y + 7y = 180^{\circ} \Rightarrow 12y = 180^{\circ} \Rightarrow y = 15^{\circ}$.
Thus,$\angle B = 5(15^{\circ}) = 75^{\circ}$ and $\angle D = 7(15^{\circ}) = 105^{\circ}$.
Therefore,the angles are $\angle A = 80^{\circ}, \angle B = 75^{\circ}, \angle C = 100^{\circ}, \angle D = 105^{\circ}$.

(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle P + \angle R = 180^{\circ}$.
Given that $\angle P = \angle R + 50^{\circ}$.
Substituting the value of $\angle P$ in the first equation:
$(\angle R + 50^{\circ}) + \angle R = 180^{\circ}$
$2\angle R + 50^{\circ} = 180^{\circ}$
$2\angle R = 130^{\circ}$
$\angle R = 65^{\circ}$.
Now,finding $\angle P$:
$\angle P = 65^{\circ} + 50^{\circ} = 115^{\circ}$.
Thus,$\angle P = 115^{\circ}$ and $\angle R = 65^{\circ}$.

(B) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$ and $\angle B + \angle D = 180^{\circ}$.
Given $\angle A = 4 \angle C$,substituting this into the first equation: $4 \angle C + \angle C = 180^{\circ} \implies 5 \angle C = 180^{\circ} \implies \angle C = 36^{\circ}$.
Then,$\angle A = 4 \times 36^{\circ} = 144^{\circ}$.
Given $\angle B = 3 \angle D$,substituting this into the second equation: $3 \angle D + \angle D = 180^{\circ} \implies 4 \angle D = 180^{\circ} \implies \angle D = 45^{\circ}$.
Then,$\angle B = 3 \times 45^{\circ} = 135^{\circ}$.
Thus,the angles are $\angle A = 144^{\circ}, \angle B = 135^{\circ}, \angle C = 36^{\circ}, \angle D = 45^{\circ}$.

(A) $1$. Since $A$ is the centre of the circle passing through $B, C$,and $D$,we have $AB = AC = AD = r$ (where $r$ is the radius of the circle).
$2$. In $\triangle ABC$,since $AB = AC$,the angles opposite to these sides are equal,so $\angle ABC = \angle ACB$.
$3$. In $\triangle ADC$,since $AD = AC$,the angles opposite to these sides are equal,so $\angle ADC = \angle ACD$.
$4$. In $\triangle BCD$,the sum of angles is $180^{\circ}$. Thus,$\angle CBD + \angle CDB + \angle BCD = 180^{\circ}$.
$5$. Note that $\angle BCD = \angle ACB + \angle ACD$. Substituting this into the sum,we get $\angle CBD + \angle CDB + \angle ACB + \angle ACD = 180^{\circ}$.
$6$. Since $\angle CBD = \angle ACB$ and $\angle CDB = \angle ACD$ is not necessarily true,let's use the property of the central angle. The angle subtended by arc $CD$ at the centre is $\angle CAD$ and at the circumference is $\angle CBD$. Thus,$\angle CAD = 2 \angle CBD$ is only true if $B$ is on the major arc.
$7$. Correct approach: In $\triangle ABD$,$AB=AD$,so $\angle ABD = \angle ADB = (180^{\circ} - \angle BAD)/2 = 90^{\circ} - \frac{1}{2} \angle BAD$.
$8$. In $\triangle BCD$,$\angle CBD + \angle CDB = 180^{\circ} - \angle BCD$. Since $A$ is the centre,$\angle BCD = \frac{1}{2} (360^{\circ} - \angle BAD) = 180^{\circ} - \frac{1}{2} \angle BAD$ (angle at circumference is half the reflex angle at centre).
$9$. Therefore,$\angle CBD + \angle CDB = 180^{\circ} - (180^{\circ} - \frac{1}{2} \angle BAD) = \frac{1}{2} \angle BAD$.

(N/A) $1$. In $\triangle ABC$,$BM \perp AC$ and $CN \perp AB$.
$2$. Therefore,$\angle BMC = 90^\circ$ and $\angle BNC = 90^\circ$.
$3$. Consider the line segment $BC$ as a chord.
$4$. The angles subtended by the line segment $BC$ at points $M$ and $N$ are $\angle BMC = 90^\circ$ and $\angle BNC = 90^\circ$.
$5$. Since $\angle BMC = \angle BNC = 90^\circ$,the points $M$ and $N$ lie on a circle with $BC$ as the diameter.
$6$. Thus,the points $B, C, M$ and $N$ are concyclic.

(N/A) Let $ABCD$ be a trapezium such that $AB \parallel CD$ and the non-parallel sides $AD = BC$.
To prove that $ABCD$ is a cyclic quadrilateral,we need to show that the sum of opposite angles is $180^{\circ}$,i.e.,$\angle A + \angle C = 180^{\circ}$ or $\angle B + \angle D = 180^{\circ}$.
Draw $DE \perp AB$ and $CF \perp AB$.
In $\triangle ADE$ and $\triangle BCF$:
$AD = BC$ (Given)
$\angle AED = \angle BFC = 90^{\circ}$ (By construction)
$DE = CF$ (Distance between parallel lines is constant)
Therefore,$\triangle ADE \cong \triangle BCF$ by $RHS$ congruence rule.
This implies $\angle A = \angle B$ (by $CPCT$).
Since $AB \parallel CD$,the sum of consecutive interior angles is $180^{\circ}$,so $\angle A + \angle D = 180^{\circ}$.
Substituting $\angle B$ for $\angle A$,we get $\angle B + \angle D = 180^{\circ}$.
Since the sum of opposite angles is $180^{\circ}$,$ABCD$ is a cyclic quadrilateral.

(D) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Given that $ABCD$ is a cyclic quadrilateral,therefore $\angle A + \angle C = 180^{\circ}$.
It is also given that $\angle A = \angle C$.
Substituting $\angle A$ for $\angle C$ in the first equation,we get $\angle A + \angle A = 180^{\circ}$.
$2\angle A = 180^{\circ}$.
Therefore,$\angle A = 180^{\circ} / 2 = 90^{\circ}$.

(B) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle Q + \angle S = 180^{\circ}$.
Given that $5 \angle Q = 7 \angle S$,we can express $\angle S$ as $\angle S = \frac{5}{7} \angle Q$.
Substituting this into the sum equation: $\angle Q + \frac{5}{7} \angle Q = 180^{\circ}$.
$\frac{7 \angle Q + 5 \angle Q}{7} = 180^{\circ}$.
$\frac{12 \angle Q}{7} = 180^{\circ}$.
$\angle Q = \frac{180^{\circ} \times 7}{12}$.
$\angle Q = 15^{\circ} \times 7 = 105^{\circ}$.

(C) In a cyclic quadrilateral,angles subtended by the same arc at the circumference are equal.
Since $ABCD$ is a cyclic quadrilateral,the angles subtended by the arc $AB$ at the circumference are $\angle ADB$ and $\angle ACB$.
Therefore,$\angle ACB = \angle ADB = 55^{\circ}$.
Now,consider the triangle $\triangle ABC$. The sum of the angles in a triangle is $180^{\circ}$.
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$.
Substituting the given values: $45^{\circ} + \angle ABC + 55^{\circ} = 180^{\circ}$.
$100^{\circ} + \angle ABC = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

(D) In a cyclic quadrilateral,the sum of the opposite angles is $180^{\circ}$.
Given that $ABCD$ is a cyclic quadrilateral,the opposite angles are $\angle A$ and $\angle C$,and $\angle B$ and $\angle D$.
Therefore,$\angle B + \angle D = 180^{\circ}$.
Given $\angle B = 75^{\circ}$,we have $75^{\circ} + \angle D = 180^{\circ}$.
$\angle D = 180^{\circ} - 75^{\circ} = 105^{\circ}$.

(D) property of a cyclic quadrilateral is that the sum of its opposite angles is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$.
Substituting the given values: $(2x - 10^{\circ}) + (3x - 35^{\circ}) = 180^{\circ}$.
Combining like terms: $5x - 45^{\circ} = 180^{\circ}$.
Adding $45^{\circ}$ to both sides: $5x = 225^{\circ}$.
Dividing by $5$: $x = 45^{\circ}$.
Now,calculate $\angle A$: $\angle A = 2x - 10^{\circ} = 2(45^{\circ}) - 10^{\circ} = 90^{\circ} - 10^{\circ} = 80^{\circ}$.

(C) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$.
From the given equation,$5 \angle A = 13 \angle C$,we can write $\angle C = \frac{5}{13} \angle A$.
Substituting this into the sum equation: $\angle A + \frac{5}{13} \angle A = 180^{\circ}$.
$\frac{13 \angle A + 5 \angle A}{13} = 180^{\circ}$.
$\frac{18 \angle A}{13} = 180^{\circ}$.
$\angle A = 180^{\circ} \times \frac{13}{18}$.
$\angle A = 10^{\circ} \times 13 = 130^{\circ}$.

(D) In a cyclic quadrilateral,angles subtended by the same arc at the circumference are equal.
Here,$\angle PRQ$ and $\angle PSQ$ are angles subtended by the same arc $PQ$ at the circumference of the circle.
Therefore,$\angle PSQ = \angle PRQ$.
Given that $\angle PRQ = 52^{\circ}$,it follows that $\angle PSQ = 52^{\circ}$.

(A) Let the angles be $3x, 2x, 4x,$ and $5x$.
Since the sum of angles in a quadrilateral is $360^{\circ}$,we have:
$3x + 2x + 4x + 5x = 360^{\circ}$
$14x = 360^{\circ}$
$x = \frac{360}{14} = \frac{180}{7}^{\circ}$
Now,calculate the angles:
$\angle A = 3 \times \frac{180}{7} = \frac{540}{7}^{\circ} \approx 77.14^{\circ}$
$\angle B = 2 \times \frac{180}{7} = \frac{360}{7}^{\circ} \approx 51.43^{\circ}$
$\angle C = 4 \times \frac{180}{7} = \frac{720}{7}^{\circ} \approx 102.86^{\circ}$
$\angle D = 5 \times \frac{180}{7} = \frac{900}{7}^{\circ} \approx 128.57^{\circ}$
For a quadrilateral to be cyclic,the sum of opposite angles must be $180^{\circ}$.
$\angle A + \angle C = \frac{540}{7} + \frac{720}{7} = \frac{1260}{7} = 180^{\circ}$
$\angle B + \angle D = \frac{360}{7} + \frac{900}{7} = \frac{1260}{7} = 180^{\circ}$
Since the sum of opposite angles is $180^{\circ}$,$ABCD$ is a cyclic quadrilateral.

(B) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$.
Substituting the given values: $(3x + 15^{\circ}) + (2x + 15^{\circ}) = 180^{\circ}$.
Combining like terms: $5x + 30^{\circ} = 180^{\circ}$.
Subtracting $30^{\circ}$ from both sides: $5x = 150^{\circ}$.
Dividing by $5$: $x = 30^{\circ}$.
Thus,the value of $x$ is $30$.

(C) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle B + \angle D = 180^{\circ}$.
Given that $\angle B = \angle D - 30^{\circ}$,we can express $\angle D$ as $\angle D = \angle B + 30^{\circ}$.
Substituting this into the sum equation: $\angle B + (\angle B + 30^{\circ}) = 180^{\circ}$.
$2\angle B + 30^{\circ} = 180^{\circ}$.
$2\angle B = 150^{\circ}$.
$\angle B = 75^{\circ}$.