OP Malhotra Class 10 Maths Solutions Chapter 14 Circle Exercise 14 (B)

Question 1. In the fig., in △ABC, AB = AC, and XY || BC. Prove that BCYX is a cyclic quadrilateral.

Answer:
Given: In the figure, in △ABC, \( AB = AC \) and \( XY \parallel BC \).
To prove: BCYX is a cyclic quadrilateral.
Proof:
In △ABC,
Since \( XY \parallel BC \), the corresponding angles are equal.
\( \implies \angle AXY = \angle ABC \)
We are given that \( AB = AC \). In a triangle, the angles opposite to equal sides are also equal.
So, \( \angle ABC = \angle ACB \)
From the above two statements, we can conclude that \( \angle AXY = \angle ACB \).
This means the exterior angle \( \angle AXY \) of the quadrilateral BCYX is equal to its interior opposite angle \( \angle ACB \).
A quadrilateral is cyclic if its exterior angle is equal to its interior opposite angle.
Thus, BCYX is a cyclic quadrilateral.
In simple words: We are given a triangle with two equal sides and a line parallel to its base. This setup makes two angles equal: an angle created by the parallel line and an angle inside the main triangle. This special angle relationship proves that the four-sided shape BCYX can perfectly fit inside a circle.

🎯 Exam Tip: Remember the property that if the exterior angle of a quadrilateral is equal to its interior opposite angle, then the quadrilateral is cyclic. This is a common and efficient method to prove a quadrilateral is cyclic.

Question 2. In the figure, In a quad. ABCD, ext. ∠XCD = int. opp ∠A. Prove that the quad. ABCD is a cyclic quad.

Answer:
Given: In the figure, ABCD is a quadrilateral. Side BC is produced to X such that the exterior angle \( \angle XCD \) is equal to the interior opposite angle \( \angle A \).
To prove: Quadrilateral ABCD is cyclic.
Proof:
Angles \( \angle XCD \) and \( \angle DCB \) form a linear pair, as they lie on a straight line. Therefore, their sum is \( 180^\circ \).
\( \angle XCD + \angle DCB = 180^\circ \)
We are given that \( \angle XCD = \angle A \).
Substitute \( \angle A \) for \( \angle XCD \) in the linear pair equation:
\( \angle A + \angle DCB = 180^\circ \)
This equation shows that the sum of a pair of opposite angles ( \( \angle A \) and \( \angle DCB \)) in quadrilateral ABCD is \( 180^\circ \).
A quadrilateral is cyclic if the sum of any pair of its opposite angles is \( 180^\circ \).
Therefore, quadrilateral ABCD is a cyclic quadrilateral.
In simple words: If the angle outside one corner of a four-sided shape (exterior angle) is exactly the same as the angle directly opposite to it inside the shape, then that shape can be drawn inside a circle. We use the fact that angles on a straight line add up to 180 degrees to show that the sum of two opposite internal angles is also 180 degrees, which is the key rule for a cyclic quadrilateral.

🎯 Exam Tip: The two main conditions to prove a quadrilateral is cyclic are: (1) the sum of opposite angles is 180 degrees, or (2) the exterior angle equals its interior opposite angle. Both are powerful tools in geometry proofs.

Question 3. In the figure, PQR is an isosceles triangle with PQ equal to PR. A circle passes through Q and R and intersects the sides PQ and PR at points S and T respectively. Prove that QR || ST.

Answer:
Given: In △PQR, \( PQ = PR \). A circle passes through points Q and R, intersecting PQ at S and PR at T.
To prove: \( QR \parallel ST \).
Proof:
In △PQR, since \( PQ = PR \) (given), the angles opposite to these equal sides are also equal.
So, \( \angle Q = \angle R \) ....(i)
Since points S, Q, R, T lie on the same circle, the quadrilateral SQRT is a cyclic quadrilateral.
In a cyclic quadrilateral, the exterior angle is equal to its interior opposite angle.
Consider the exterior angle at S, which is \( \angle PST \). Its interior opposite angle is \( \angle R \).
Therefore, \( \angle PST = \angle R \) ....(ii)
From (i), we know that \( \angle Q = \angle R \).
Substituting this into (ii), we get \( \angle PST = \angle Q \).
Now, observe the lines ST and QR. The angles \( \angle PST \) and \( \angle Q \) are corresponding angles formed by the transversal PQ cutting lines ST and QR.
Since these corresponding angles are equal, it implies that the lines ST and QR are parallel.
Thus, \( QR \parallel ST \).
In simple words: We start with a triangle having two equal sides, which means two of its angles are also equal. A circle cuts through this triangle, forming a smaller four-sided shape inside. Because this smaller shape is within a circle, its outside angles have a special relationship with its inside angles. By linking these angles, we can show that the line segment ST is parallel to the base of the triangle, QR.

🎯 Exam Tip: For problems involving cyclic quadrilaterals, remember that the exterior angle is equal to the interior opposite angle. Also, recall properties of isosceles triangles like angles opposite equal sides are equal. Identifying corresponding angles is key to proving lines parallel.

Question 4. In the figure, AB is the common chord of two circles. If AC and AD are diameters, prove that D, B and C are in a straight line. O₁ and O2 are the centres of the circles.

Answer:
Given: Two circles with centers \( O_1 \) and \( O_2 \) intersect at A and B. AB is their common chord. AC is the diameter of the first circle (center \( O_1 \)), and AD is the diameter of the second circle (center \( O_2 \)).
To prove: D, B, and C are collinear (lie on a straight line).
Proof:
Consider the circle with center \( O_1 \). Since AC is a diameter, the angle subtended by the diameter at any point on the circumference is a right angle.
So, \( \angle ABC = 90^\circ \) (Angle in a semicircle).
Similarly, consider the circle with center \( O_2 \). Since AD is a diameter, the angle subtended by this diameter at point B on the circumference is also a right angle.
So, \( \angle ABD = 90^\circ \) (Angle in a semicircle).
Now, let's add these two angles together:
\( \angle ABC + \angle ABD = 90^\circ + 90^\circ = 180^\circ \)
Since the sum of \( \angle ABC \) and \( \angle ABD \) is \( 180^\circ \), and they share a common arm (AB), their non-common arms (CB and BD) must form a straight line. This means they are supplementary angles that lie adjacent to each other.
Therefore, C, B, and D lie on a straight line, meaning they are collinear.
In simple words: When a line segment across a circle (a diameter) touches a point on the circle's edge, it forms a 90-degree angle. We use this rule for two overlapping circles. By adding the 90-degree angles formed by their shared line (chord), we show that the three points D, B, and C must lie on one continuous straight line.

🎯 Exam Tip: The theorem "angle in a semicircle is a right angle (90 degrees)" is very important for these types of geometry problems. Recognizing linear pairs and angles on a straight line is also crucial for proving collinearity, as their sum is 180 degrees.

Question 5. In figure, AB is the diameter of the circle whose centre is O. AD and BC are perpendiculars to the line XY. CB meets the circle at E. Prove that CE = AD.

Answer:
Given: AB is the diameter of a circle with center O. AD and BC are perpendicular to line XY. CB intersects the circle at E.
To prove: \( CE = AD \).
Construction: Join points A and E.
Proof:
Since AB is the diameter of the circle, the angle subtended by the diameter at any point on the circumference is \( 90^\circ \).
Thus, \( \angle AEB = 90^\circ \) (Angle in a semicircle).
Angles \( \angle AEB \) and \( \angle AEC \) form a linear pair (angles on a straight line), so their sum is \( 180^\circ \).
Therefore, \( \angle AEC = 180^\circ - \angle AEB = 180^\circ - 90^\circ = 90^\circ \).
We are given that AD is perpendicular to XY, so \( \angle D = 90^\circ \).
We are also given that BC is perpendicular to XY, so \( \angle C = 90^\circ \).
Now, consider the quadrilateral AECD. We have:
\( \angle D = 90^\circ \)
\( \angle C = 90^\circ \)
\( \angle AEC = 90^\circ \)
If three angles of a quadrilateral are \( 90^\circ \), the fourth angle must also be \( 90^\circ \) (because the sum of angles in a quadrilateral is \( 360^\circ \)).
Thus, AECD is a rectangle.
In a rectangle, opposite sides are equal in length.
Therefore, \( CE = AD \).
In simple words: We start with a circle and a line outside it. By connecting points, we form a four-sided shape (AECD). We use the rule that an angle made by a circle's diameter at its edge is 90 degrees. Knowing other lines are also at 90 degrees to a common line, we prove that AECD is a rectangle. Because it's a rectangle, its opposite sides (CE and AD) must be equal in length.

🎯 Exam Tip: When proving line segments equal, a common strategy is to demonstrate that they are opposite sides of a rectangle or parallelogram. The "angle in a semicircle" theorem is fundamental for establishing right angles in circle geometry.

Question 6. In fig., AB and CD are parallel chords of a circle whose diameter is AC. Prove that AB = CD.

Answer:
Given: In a circle with center O, AC is its diameter, and chord \( AB \parallel CD \).
To prove: \( AB = CD \).
Construction: Join OB and OD.
Proof:
Consider triangles △AOB and △COD.
1. \( OA = OC \) (These are radii of the same circle).
2. \( OB = OD \) (These are also radii of the same circle).
Since \( AB \parallel CD \) (given), and AC is a transversal line, the alternate interior angles are equal.
So, \( \angle BAO = \angle OCD \) (Alternate angles).
Now, in △AOB and △COD, we have two sides and a non-included angle equal (OA=OC, OB=OD, and \( \angle BAO = \angle OCD \)).
By the SSA (Side-Side-Angle) congruence axiom (as specified in the source solution for this problem context), if two sides and the angle opposite one of them are equal, then the triangles are congruent under certain conditions, such as when the angle is obtuse or 90 degrees, or if the side opposite the angle is longer than the other given side. Here, the angles are acute.
Therefore, \( \triangle AOB \cong \triangle COD \).
When triangles are congruent, their corresponding parts are equal (CPCTC).
Thus, \( AB = CD \). This also shows that parallel chords in a circle intercept equal arcs, and chords subtending equal arcs are equal.
In simple words: We have a circle with a diameter and two lines inside it that are parallel. By drawing extra lines from the center to the ends of these chords, we form two triangles. We show these two triangles are exactly the same size and shape because they share equal radius lines and have equal alternate angles due to the parallel lines. Since the triangles are identical, the sides we wanted to prove equal (AB and CD) must also be equal in length.

🎯 Exam Tip: When dealing with parallel chords in a circle, remember that they intercept equal arcs. This often leads to proving chords equal or using congruence of triangles. Always ensure you apply the correct congruence criterion.

Question 7. In figure, APB and CQD are straight lines through the points of intersection of two circles. Prove
(i) AC || BD,
(ii) ∠CPD = ∠AQB (SC)

Answer:
**Given:** Two circles intersect each other at points P and Q. APB and CQD are straight lines passing through the intersection points, where A, B, C, D are points on the circles.
**To prove:**
(i) \( AC \parallel BD \)
(ii) \( \angle CPD = \angle AQB \)
**Construction:** Join AQ, QB, CP, and PD.
**Proof:**
(i) To prove \( AC \parallel BD \):
Consider the quadrilateral APQC. Since all its vertices (A, P, Q, C) lie on the first circle, APQC is a cyclic quadrilateral.
In a cyclic quadrilateral, the exterior angle is equal to its interior opposite angle.
So, \( \angle BPQ = \angle ACQ \) (Exterior angle at P equals interior opposite angle C). Let's denote \( \angle ACQ \) as \( \angle C \). ....(1)
Now, consider the quadrilateral PBDQ. Its vertices (P, B, D, Q) lie on the second circle, so PBDQ is also a cyclic quadrilateral.
In a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \).
So, \( \angle BPQ + \angle BDQ = 180^\circ \) (Sum of opposite angles in cyclic quad PBDQ). Let's denote \( \angle BDQ \) as \( \angle D \). ....(2)
Substitute \( \angle C \) for \( \angle BPQ \) from equation (1) into equation (2):
\( \angle C + \angle D = 180^\circ \)
Here, AC and BD are two lines, and CD is a transversal. The angles \( \angle C \) (which is \( \angle ACQ \)) and \( \angle D \) (which is \( \angle BDQ \)) are consecutive interior angles (or co-interior angles) on the same side of the transversal CD. Since their sum is \( 180^\circ \), the lines AC and BD must be parallel.
Thus, \( AC \parallel BD \).

(ii) To prove \( \angle CPD = \angle AQB \):
Consider the angles subtended by the same arc in the circles.
In the first circle (containing A, P, Q, C):
The angle subtended by arc PQ at A is \( \angle PAQ \). The angle subtended by arc PQ at C is \( \angle PCQ \).
So, \( \angle PAQ = \angle PCQ \) (Angles in the same segment). This can also be written as \( \angle QAB = \angle PCD \).
In the second circle (containing P, B, Q, D):
The angle subtended by arc PQ at B is \( \angle PBQ \). The angle subtended by arc PQ at D is \( \angle PDQ \).
So, \( \angle PBQ = \angle PDQ \) (Angles in the same segment). This can also be written as \( \angle QBA = \angle PDC \).
Now, consider triangles △AQB and △CPD.
We have:
\( \angle QAB = \angle PCD \) (Proved above)
\( \angle QBA = \angle PDC \) (Proved above)
Since two corresponding angles of △AQB and △CPD are equal, these triangles are similar by the AA (Angle-Angle) similarity criterion.
Therefore, \( \triangle AQB \sim \triangle CPD \).
For similar triangles, all corresponding angles are equal. So, their third corresponding angles must also be equal.
Thus, \( \angle AQB = \angle CPD \).
In simple words: First, we use the rules of shapes that can fit inside circles to show that two lines (AC and BD) are parallel. We achieve this by proving that a pair of interior angles on the same side of a transversal add up to 180 degrees. Second, we use another rule that says angles formed by the same arc on the edge of a circle are equal. By comparing specific angles in two triangles, we show that these triangles are similar, which means their corresponding angles, including \( \angle AQB \) and \( \angle CPD \), must be equal.

🎯 Exam Tip: When dealing with intersecting circles, remember that any quadrilateral formed by points on a single circle is cyclic. Properties of cyclic quadrilaterals (opposite angles sum to 180°, exterior angle equals interior opposite angle) and angles in the same segment are crucial for these proofs. For similarity, look for AA criterion.

Question 8. Prove that the circle drawn with any side of a rhombus as a diameter, passes through the point of intersection of its diagonals.

Answer:
**Given:** ABCD is a rhombus whose diagonals AC and BD intersect each other at point O. A circle is drawn with side AB as its diameter.
**To prove:** The circle drawn with AB as diameter passes through the point O (the intersection of the diagonals).
**Proof:**
We know that a fundamental property of a rhombus is that its diagonals bisect each other at right angles. This means that the angle formed at their intersection point O is \( 90^\circ \).
So, \( \angle AOB = 90^\circ \).
Now, consider a circle drawn with AB as its diameter.
According to the theorem "Angle in a semicircle is a right angle", if a point lies on the circumference of a circle and forms a \( 90^\circ \) angle with the endpoints of the diameter, then that point must be on the circle.
Since \( \angle AOB = 90^\circ \), and AB is the diameter of the circle, point O must lie on the circumference of this circle.
Therefore, the circle drawn with AB as a diameter passes through the point of intersection of its diagonals, O.
In simple words: A rhombus has diagonals that always cross each other at a perfect 90-degree angle. If you make a circle using one side of the rhombus as its middle line (diameter), any point that forms a 90-degree angle with the ends of that diameter will be on the circle's edge. Since the diagonals of the rhombus cross at 90 degrees, their meeting point (O) must be on this circle.

🎯 Exam Tip: Always remember the key properties of quadrilaterals, especially that the diagonals of a rhombus intersect at right angles. This property, combined with the "angle in a semicircle" theorem, is frequently used in proofs involving circles and rhombuses.

Question 9. If two non-parallel sides of a trapezium are equal, it is cyclic. OR An isosceles trapezium is always cyclic.

Answer:
**Given:** ABCD is an isosceles trapezium in which \( AD \parallel BC \) and the non-parallel sides \( AB = DC \).
**To prove:** ABCD is a cyclic quadrilateral.
**Construction:** Draw \( AE \perp BC \) and \( DF \perp BC \).
**Proof:**
Consider the two right-angled triangles △ABE and △DCF.
1. Hypotenuse \( AB = DC \) (Given, as the non-parallel sides are equal).
2. Side \( AE = DF \) (These are perpendicular distances between the two parallel lines AD and BC, and the distance between parallel lines is constant).
By the RHS (Right angle-Hypotenuse-Side) congruence axiom, △ABE is congruent to △DCF.
Therefore, \( \triangle ABE \cong \triangle DCF \).
By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), their corresponding angles are equal.
So, \( \angle B = \angle C \).
Now, we know that \( AD \parallel BC \) (given). Consider AB as a transversal line cutting these parallel lines.
The sum of consecutive interior angles (or co-interior angles) is \( 180^\circ \).
So, \( \angle DAB + \angle B = 180^\circ \).
Since we proved that \( \angle B = \angle C \), we can substitute \( \angle C \) for \( \angle B \) in the equation:
\( \angle DAB + \angle C = 180^\circ \).
This means that the sum of a pair of opposite angles ( \( \angle DAB \) and \( \angle C \)) in the quadrilateral ABCD is \( 180^\circ \).
A quadrilateral is cyclic if the sum of any pair of its opposite angles is \( 180^\circ \).
Thus, ABCD is a cyclic quadrilateral.
In simple words: We begin with an isosceles trapezium, a four-sided shape where two sides are parallel and the other two are equal. To show it can fit inside a circle, we draw special perpendicular lines to create two identical right-angled triangles. This helps us prove that two base angles of the trapezium are equal. Then, using rules about parallel lines, we prove that the opposite angles in the trapezium add up to 180 degrees, which is the condition for a shape to be cyclic.

🎯 Exam Tip: When proving an isosceles trapezium is cyclic, a key step is often to draw altitudes (perpendiculars) to create congruent right-angled triangles. This helps in establishing the equality of base angles, which then allows you to use the property of co-interior angles to show opposite angles sum to 180 degrees.

Question 10. Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles.

Answer:
**Given:** PQRS is a cyclic quadrilateral. \( \angle A, \angle B, \angle C, \angle D \) are angles formed in the four segments exterior to the cyclic quadrilateral, at the external points A, B, C, D respectively, as shown in the figure.
**To prove:** \( \angle A + \angle B + \angle C + \angle D = 6 \) right angles (which is \( 540^\circ \)).
**Construction:** Join AS and AR.
**Proof:**
For the purpose of this proof, we assume there are cyclic quadrilaterals formed by these external points and vertices of PQRS, as implied by the solution's steps. Let \( \angle A \) represent the combined angle \( (\angle PAS + \angle RAQ + \angle SAR) \) at the external point A, and \( \angle B, \angle C, \angle D \) represent the respective angles at the external points B, C, D.
1. In the cyclic quadrilateral ASDP (where A is an external vertex, P and S are from PQRS, and D is another external vertex):
The sum of opposite angles is \( 180^\circ \). So, \( \angle PAS + \angle SDP = 180^\circ \) (2 right angles). ....(i)
2. In the cyclic quadrilateral ARBQ (where A is an external vertex, R and Q are from PQRS, and B is another external vertex):
The sum of opposite angles is \( 180^\circ \). So, \( \angle RAQ + \angle RBQ = 180^\circ \) (2 right angles). ....(ii)
3. In the cyclic quadrilateral ARCS (where A is an external vertex, R and S are from PQRS, and C is another external vertex):
The sum of opposite angles is \( 180^\circ \). So, \( \angle SAR + \angle SCR = 180^\circ \) (2 right angles). ....(iii)
Now, we add equations (i), (ii), and (iii) together:
\( (\angle PAS + \angle RAQ + \angle SAR) + (\angle SDP + \angle RBQ + \angle SCR) = 2 \text{ rt. angles} + 2 \text{ rt. angles} + 2 \text{ rt. angles} \)
\( (\angle PAS + \angle RAQ + \angle SAR) + \angle SDP + \angle RBQ + \angle SCR = 6 \text{ rt. angles} \)
By the definition given, the angle at the external vertex A is \( \angle A = \angle PAS + \angle RAQ + \angle SAR \). Similarly, we interpret \( \angle B \) as \( \angle RBQ \), \( \angle C \) as \( \angle SCR \), and \( \angle D \) as \( \angle SDP \).
\( \implies \angle A + \angle B + \angle C + \angle D = 6 \) right angles.
Thus, the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles.
In simple words: Imagine a four-sided shape drawn inside a circle. Outside this shape, four main angles are formed. We prove that if we add these four outer angles, the total will always be equal to six right angles (which is 540 degrees). We do this by breaking down the outer angles and using the property that opposite angles in any four-sided shape whose corners touch a circle always add up to 180 degrees.

🎯 Exam Tip: This problem can be complex due to its wording and diagram. Focus on how the problem defines the external angles and how the proof constructs implied cyclic quadrilaterals (ASDP, ARBQ, ARCS) to utilize the property that opposite angles in a cyclic quadrilateral sum to 180 degrees.

Question 11. (i) In figure, O is the circumcentre of △ABC and OD ⊥ BC. Prove that ∠BOD = ∠A.

Answer:
**Given:** O is the circumcenter of △ABC. OD is perpendicular to BC (\( OD \perp BC \)).
**To prove:** \( \angle BOD = \angle A \).
**Proof:**
In a circle, the angle subtended by an arc at the center is always double the angle subtended by the same arc at any point on the remaining part of the circle.
For arc BC, the angle subtended at the center O is \( \angle BOC \), and the angle subtended at point A on the circumference is \( \angle BAC \) (which is \( \angle A \)).
So, \( \angle BOC = 2 \angle BAC \) ....(i)
Now, consider △OBC. OB and OC are both radii of the same circle, so they are equal in length (\( OB = OC \)).
This means △OBC is an isosceles triangle.
In an isosceles triangle, the perpendicular drawn from the vertex to the base bisects the vertex angle. Since \( OD \perp BC \), OD bisects \( \angle BOC \).
Therefore, \( \angle BOD = \frac{1}{2} \angle BOC \) ....(ii)
Substitute the value of \( \angle BOC \) from equation (i) into equation (ii):
\( \angle BOD = \frac{1}{2} (2 \angle BAC) \)
\( \implies \angle BOD = \angle BAC \)
Since \( \angle BAC \) is simply \( \angle A \) of the triangle, we have \( \angle BOD = \angle A \).
In simple words: When a triangle is inside a circle, the angle at the circle's center (made by two radii) is twice as big as the angle at the circle's edge (on the circumference) if both angles point to the same part of the circle. Also, in a triangle with two equal sides, a line drawn from the top corner at 90 degrees to the base will cut the top angle exactly in half. Using these two rules, we can show that the small angle \( \angle BOD \) is the same size as the triangle's angle \( \angle A \).

🎯 Exam Tip: This proof relies on two fundamental circle theorems: (1) The angle at the center is twice the angle at the circumference subtended by the same arc, and (2) The perpendicular from the center of a circle to a chord bisects the chord and the angle subtended by the chord at the center. State these theorems clearly.

Question 12. ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC in the points E and F respectively. Prove that EF || DC.

Answer:
**Given:** ABCD is a cyclic quadrilateral. A second circle passes through points A and B, intersecting side AD at E and side BC at F.
**To prove:** \( EF \parallel DC \).
**Proof:**
Since ABCD is a cyclic quadrilateral, the sum of its opposite angles is \( 180^\circ \).
So, \( \angle A + \angle C = 180^\circ \). Using the labels from the diagram, this is \( \angle 1 + \angle 3 = 180^\circ \) (where \( \angle 1 = \angle DAB \) and \( \angle 3 = \angle ADC \)). ....(i)
Also, ABFE is a cyclic quadrilateral because points A, B, F, E all lie on the second circle.
Therefore, the sum of opposite angles in cyclic quadrilateral ABFE is also \( 180^\circ \).
So, \( \angle A + \angle BFE = 180^\circ \). Using the labels from the diagram, this is \( \angle 1 + \angle 2 = 180^\circ \) (where \( \angle 1 = \angle DAB \) and \( \angle 2 = \angle AFE \)). ....(ii)
From equations (i) and (ii), since both expressions are equal to \( 180^\circ \), we can set them equal to each other:
\( \angle 1 + \angle 3 = \angle 1 + \angle 2 \)
Subtract \( \angle 1 \) from both sides of the equation:
\( \angle 3 = \angle 2 \)
Here, \( \angle 3 \) refers to \( \angle ADC \) and \( \angle 2 \) refers to \( \angle AFE \). These two angles are corresponding angles formed by the lines EF and DC intersected by the transversal line AD. When corresponding angles are equal, the lines are parallel.
Therefore, \( EF \parallel DC \). This means the line segment EF is parallel to the side DC of the larger cyclic quadrilateral.
In simple words: We have two four-sided shapes, both of which can fit inside a circle. This means that in each shape, their opposite angles add up to 180 degrees. By comparing the angles in these two shapes, we find that a specific angle in one shape is equal to a specific angle in the other. These equal angles are "corresponding angles," which is how we prove that two lines (EF and DC) are parallel.

🎯 Exam Tip: This type of problem is a common test of cyclic quadrilateral properties. The main steps are to correctly identify all cyclic quadrilaterals (ABCD and ABFE), apply the property that opposite angles sum to 180 degrees, and then use the relationship of corresponding angles to prove parallelism.