Ch 5 Solutions: I’m Up and Down, and Round and Round Class 9 Math NCERT

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Exercise Set 5.1

Q1. Draw ΔABC with AB = 5 cm, ∠A = 70°, and ∠B = 60°. Draw the
circumcircle of ΔABC. Is the centre inside or outside the triangle?

Solution:

This question is based on Section 5.3 (Page 96–97) of your textbook. The key idea is:
“There is a unique circle passing through three non-collinear points.” — Theorem 1, Page 96
The centre of this circle is called the circumcentre, and the circle is called the circumcircle. To find the circumcentre, you draw the perpendicular bisectors of the sides of the triangle — they all meet at one point.

Step 1 — Find the Missing Angle ∠C

Using the angle sum property of triangles.

∠A + ∠B + ∠C = 180°

70° + 60° + ∠C = 180°

∠C = 180° − 130° = 50°

All three angles — 70°, 60°, and 50° — are less than 90°. This means ΔABC is an acute-angled triangle.

Step 2 — Construct Triangle ABC

1. Draw a line segment AB = 5 cm on your page. Mark point A on the left and point B on the right.
2. Place your protractor at point A and mark an angle of 70° above line AB. Draw a ray from A in that direction.
3. Place your protractor at point B and mark an angle of 60° above line AB. Draw a ray from B in that direction.
4. The two rays will cross each other at a point — label this point C.
5. Your triangle ABC is now complete, with AB = 5 cm, ∠A = 70°, ∠B = 60°, and ∠C = 50°.

Step 3 — Draw the Perpendicular Bisectors

The circumcentre O is found where the perpendicular bisectors of the sides meet. You only need two bisectors — the third will automatically pass through the same point.

For side AB:

1. Open your compass to more than half the length of AB.
2. Place the compass tip at A and draw arcs above and below AB.
3. Without changing the compass width, place the tip at B and draw arcs that cross the first arcs.
4. Connect the two crossing points with a line. This is the perpendicular bisector of AB.

For side BC:

1. Repeat the same process with the compass tip at B and then at C.
2. Draw the perpendicular bisector of BC.

Mark the point where these two bisectors cross as O. This is your circumcentre.

Step 4 — Draw the Circumcircle

1. Place the compass tip at O.
2. Open the compass until the pencil touches point A.
3. Draw a full circle.

The circle will pass exactly through A, B, and C. This is the circumcircle of ΔABC. You can verify by checking that the compass also reaches B and C without adjusting — since OA = OB = OC (all equal to the circumradius).

Is the circumcentre O inside or outside the triangle?
Answer: The circumcentre O lies INSIDE the triangle.
This is confirmed by the rule stated on Page 97 of your textbook:
“For an acute-angled triangle, the circumcentre lies inside the triangle.”
Since all three angles of ΔABC are less than 90°, the triangle is acute. So the circumcentre will always fall inside — and your construction will show exactly that.

Q2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the
circumcircle of ΔABC. Is the centre inside or outside the triangle?

Solution:

Q3. Draw ΔABC, with AB = 6 cm, BC = 7 cm, and CA = 7 cm. Draw
the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.

Solution:

Q4. What is the least possible radius of a circle through two points A and B?

Solution:

Exercise Set 5.2

Q1. Show that the triangle formed by a chord and the centre of the
circle is isosceles.

Solution:

Q2. Show that if two such isosceles triangles (occurring in the previous question)
have equal base length, they are congruent to each other.

Solution:

Exercise Set 5.3

Q1. Can you explain why the converse to Theorem 4 is true, i.e., why
does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?
(Hint: Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)

Solution:

Q2. An isosceles triangle ABC is inscribed in a circle, with AB = AC.
Show that the altitude from A to BC passes through the centre of
the circle.

Solution:

Q3. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides
of the centre of a circle. If the radius of the circle is 5 cm, find the
distance between the midpoints of the chords.

Solution:

Exercise Set 5.4

Q1. Use the Baudhāyana–Pythagoras theorem to show why the Theorem
6 must be true.

Solution:

Q2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH,
and CE= CH, show that AB = GF.

Solution:

Q3. Solve the previous question using the Baudhāyana–Pythagoras theorem.

Solution:

Exercise Set 5.5

Q1. Find the length of the chord of a circle where the radius is 7 cm
and the perpendicular distance is 6 cm.

Q2. Explain why the following statement is true: If the perpendicular
distance of a chord from the centre is d and the radius is r, then the
chord length is $2\sqrt{r^2 – d^2}$

Q3. In a circle, if the distance of chord AB from the centre is twice
the distance of another chord CD from the centre, then can we
conclude that CD = 2 AB? Give reasons for your answer

Exercise Set 5.6

Q2. In a circle with centre O, the central angle AOB is 60°. If the radius
of the circle is 12 cm, what is the length of the chord AB?

Q3. Find x in Fig. 5.26.

End-of-Chapter Exercises

Q1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?

Given

• Radius of the circle $r = 13 \text{ cm}$
• Perpendicular distance of chord from centre $d = 5 \text{ cm}$

We know:

$\text{Chord Length}=2\sqrt{r^2-d^2}$

Substitute the values:$$\text{Chord Length} =2\sqrt{13^2-5^2}$$
$$=2\sqrt{169-25}$$
$$=2\sqrt{144}$$
$$=2\times 12$$
$$=24 \text{ cm}$$

24 cm Ans

Q2. An arc of a circle subtends an angle of 70° at the centre. What is
The measure of the angle subtended by the arc at a point on the circle?

Solution:

Given

The angle subtended by the arc at the centre is:$$70^\circ$$

We know the theorem:

$\angle\text{ at centre }=2\times\angle\text{ at circumference}$

Let the angle subtended at a point on the circle be $x$

Then,$$70^\circ = 2x$$

Divide both sides by 2:$$x=\frac{70^\circ}{2}$$
$$x=35^\circ$$

35. Ans

Q3. The diameter of a circle is 26 cm. A chord of length 24 cm is
drawn in the circle. Find the distance from the centre of the circle
to the chord.

Solution:

Given

• Diameter of the circle $= 26 \text{ cm}$

So, radius:$$r=\frac{26}{2}=13 \text{ cm}$$

• Length of the chord $=24 \text{ cm}$

Half of the chord:$$\frac{24}{2}=12 \text{ cm}$$

Let:

• $O$ be the centre of the circle
• $AB$ be the chord
• $M$ be the midpoint of the chord $AB$

The perpendicular from the centre to a chord bisects the chord.

Therefore,$$AM=12 \text{ cm}$$

In right triangle $OMA$
$$OA=13 \text{ cm}$$
$$AM=12 \text{ cm}$$

Using the Baudhāyana–Pythagoras theorem:

$OA^2=OM^2+AM^2$

Substituting the values:$$13^2=OM^2+12^2$$
$$169=OM^2+144$$
$$OM^2=169-144$$
$$OM^2=25$$
$$OM=5 \text{ cm}$$

5 cm. Ans

Q4. A circle has a radius of 15 cm. A chord is drawn. The distance
from the centre of the circle to the chord is 9 cm. What is the
length of the chord?

Solution:

Given

• Radius of the circle:

$$r=15 \text{ cm}$$

• Distance from the centre to the chord:

$$d=9 \text{ cm}$$

Let:

• $AB$ be the chord
• $O$ be the centre of the circle
• $M$ be the midpoint of the chord $AB$AB

The perpendicular from the centre to a chord bisects the chord.

Therefore,$$AM=\frac{AB}{2}$$

In right triangle $OMA$,$$OA=15 \text{ cm}$$
$$OM=9 \text{ cm}$$

Using the Baudhāyana–Pythagoras theorem:

$OA^2=OM^2+AM^2$

Substituting the values:$$15^2=9^2+AM^2$$
$$225=81+AM^2$$
$$AM^2=225-81$$
$$AM^2=144$$

$$AM=12 \text{ cm}$$

Since $AB=2AM$$$AB=2\times 12$$
$$AB=24 \text{ cm}$$AB = 24 cm. Ans

Q5. Prove that the perpendicular bisector of a chord passes through
the centre of the circle.

Solution:

Given

• $AB$ is a chord of a circle.
• $O$ is the centre of the circle.
• $MN$ is the perpendicular bisector of chord $AB$.

To Prove:

The perpendicular bisector of chord $AB$ passes through the centre $O$.

Construction

Join $OA$ and $OB$.

Let the perpendicular bisector meet chord $AB$ at point $M$.

Then,$$AM=MB$$

and$$OM \perp AB$$

Proof

In triangles $\triangle OMA$ and $\triangle OMB$:$$OA=OB$$

(Radii of the same circle)$$AM=MB$$

($M$ is the midpoint of $AB$)

and$$OM=OM$$

(Common side)

Therefore,$$\triangle OMA \cong \triangle OMB$$

by SSS congruence.

Hence,$$\angle OMA=\angle OMB$$

But these two angles form a linear pair.

So,$$\angle OMA=\angle OMB=90^\circ$$

Thus, $OM$ is perpendicular to chord $AB$ at its midpoint.

Therefore, the perpendicular bisector of the chord passes through the centre $O$.

Hence Proved.

Q6. The diameter of a circle is AB. Point C is on the circumference.
What is the measure of the ∠ACB? Explain your reasoning

Solution :

Given

• $AB$ is the diameter of the circle.
• $C$ is a point on the circumference.

We need to find:$$\angle ACB$$

Reasoning

The angle subtended by a diameter at the circumference is always a right angle.

This is called the theorem:

$\text{Angle in a semicircle}=90^\circ$

Since $AB$ is the diameter, the arc $AB$ forms a semicircle.

Therefore, the angle subtended by $AB$ at point $C$ is:$$\angle ACB=90^\circ$$

$$\boxed{\angle ACB=90^\circ}$$

Q7. ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A
measures 75°, what is the measure of ∠C? If ∠B measures 110°,
What is the measure of ∠D?

Solution:

Given

ABCD is a cyclic quadrilateral.

In a cyclic quadrilateral, opposite angles are supplementary.

That means:

$\angle A+\angle C=180^\circ$

and

$\angle B+\angle D=180^\circ$

Finding $\angle C$

Given:$$\angle A=75^\circ$$

So,$$75^\circ+\angle C=180^\circ$$
$$\angle C=180^\circ-75^\circ$$
$$\angle C=105^\circ$$

Finding $\angle D$

Given:$$\angle B=110^\circ$$

So,$$110^\circ+\angle D=180^\circ$$
$$\angle D=180^\circ-110^\circ$$
$$\angle D=70^\circ$$

Ans.

$$\boxed{\angle C=105^\circ}$$
$$\boxed{\angle D=70^\circ}$$

Q8. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)°
and ∠R = (3x − 20)°, find the value of x and the measures of
∠P and ∠R.

Solution:

Given

Quadrilateral $PQRS$ is a cyclic quadrilateral.$$\angle P=(2x+10)^\circ$$
$$\angle R=(3x-20)^\circ$$

In a cyclic quadrilateral, opposite angles are supplementary.

So,

$\angle P+\angle R=180^\circ$

Substituting the values:$$(2x+10)^\circ+(3x-20)^\circ=180^\circ$$
$$2x+10+3x-20=180$$
$$5x-10=180$$
$$5x=190$$
$$x=38$$

Finding $\angle P$

$$\angle P=2x+10$$
$$=2(38)+10$$
$$=76+10$$
$$=86^\circ$$

Finding $\angle R$

$$\angle R=3x-20$$
$$=3(38)-20$$
$$=114-20$$
$$=94^\circ$$

Ans.

$$\boxed{x=38}$$
$$\boxed{\angle P=86^\circ}$$
$$\boxed{\angle R=94^\circ}$$

Q9. The distance of a chord of length 16 cm from the centre of a
circle is 6 cm. Find the radius of the circle.

Solutions:

Given

• Length of the chord $=16 \text{ cm}$

Half of the chord:
$$\frac{16}{2}=8 \text{ cm}$$

• Distance of the chord from the centre:
  

$$OM=6 \text{ cm}$$

Let:

• $O$ be the centre of the circle
• $AB$ be the chord
• $M$ be the midpoint of the chord $AB$

Since the perpendicular from the centre to a chord bisects the chord,
$$AM=8 \text{ cm}$$

In right triangle $OMA$,
$$OM=6 \text{ cm}$$
$$AM=8 \text{ cm}$$

Using the Baudhāyana–Pythagoras theorem:

$OA^2=OM^2+AM^2$

Substituting the values:
$$OA^2=6^2+8^2$$
$$OA^2=36+64$$
$$OA^2=100$$
$$OA=10 \text{ cm}$$

Ans.

$$\boxed{10\text{ cm}}$$

Q10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

Solution:

Given

A cyclic quadrilateral has sides:
$$5,\ 5,\ 12,\ 12$$

Since opposite sides are equal,

The quadrilateral is a rectangle.

So,
$$\text{Length}=12$$
$$\text{Breadth}=5$$

Area of rectangle:

$\text{Area}=\text{Length}\times\text{Breadth}$

Substituting the values:
$$\text{Area}=12\times 5$$
$$\text{Area}=60$$

Ans.

$$\boxed{60\text{ square units}}$$

Q11. Consider a cyclic quadrilateral. Without drawing its circumcircle,
how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of
finding out?

Solution:

The position of the circumcentre depends on the type of angles in the cyclic quadrilateral.

The circumcentre is formed by the intersection of the perpendicular bisectors of the sides.

Best Method

Observe the angles of the cyclic quadrilateral.

Case 1: All angles are acute

If all the angles are less than $90^\circ$,

then the circumcentre lies inside the quadrilateral.

Case 2: One angle is obtuse

If one angle is greater than $90^\circ$,

then the circumcentre lies outside the quadrilateral.

Why?

In a cyclic quadrilateral:
$$\text{Opposite angles are supplementary}$$

So if one angle becomes obtuse, the geometry of the perpendicular bisectors shifts the circumcentre outside the figure.

Best Practical Way

The easiest way is:

1. Check whether the quadrilateral has an obtuse angle.
2. If yes → circumcentre lies outside.
3. If all angles are acute → circumcentre lies inside.

Conclusion

• All acute angles $\Rightarrow$⇒ circumcentre inside.
• At least one obtuse angle $\Rightarrow$⇒ circumcentre outside.

Q12. When two chords intersect, each of them is divided into two
line segments. Show that if the intersecting chords are of equal
length, then the line segments of one chord are equal to the
corresponding line segments of the other chord.

Solution:

Given

Two chords intersect inside a circle.

Let the chords $AB$ and $CD$ intersect at point $P$.

Given:
$$AB=CD$$

Suppose:$$AP=x,\quad PB=y$$

and$$CP=m,\quad PD=n$$

Since the whole chords are equal,
$$AB=CD$$

So,$$x+y=m+n$$

Also, by the intersecting chords theorem:

$AP\times PB=CP\times PD$

Therefore,
$$xy=mn$$

Now we have:
$$x+y=m+n$$

and
$$xy=mn$$

Two numbers with the same sum and the same product must be equal.

Hence,
$$x=m$$

and$$y=n$$

That is,
$$AP=CP$$

and
$$PB=PD$$

Therefore,

The line segments of one chord are equal to the corresponding line segments of the other chord.

Hence Proved.

Q13. Draw a circle in which a chord of 6 cm length stands at a
distance of 3 cm from the centre. (Hint: Is it a circumcircle of a suitable triangle?)

Solution:

Construction

A chord of length $6\text{ cm}$ stands at a distance $3\text{ cm}$ from the centre.

The perpendicular from the centre to the chord bisects the chord.

So half of the chord is:
$$\frac{6}{2}=3\text{ cm}$$

Now we get a right triangle where:

• One side $=3\text{ cm}$
• Other side $=3\text{ cm}$

Using the Baudhāyana–Pythagoras theorem:

$r^2=3^2+3^2$
$$r^2=9+9$$
$$r^2=18$$
$$r=3\sqrt2\text{ cm}$$

Steps of Construction

1. Draw a line segment $AB=6\text{ cm}$
2. Find the midpoint $M$ of $AB$
3. Draw a perpendicular line at $M$.
4. On the perpendicular line, mark a point $O$ such that:

$$OM=3\text{ cm}$$

5. Join $OA$ and $OB$.
6. With centre $O$ and radius $OA$, draw the circle.

This is the required circle

Q14. Show that a rectangle is the only parallelogram that can be
inscribed in a circle.

Solution :

Given

ABCD is a parallelogram inscribed in a circle.

To Prove:

The parallelogram must be a rectangle.

Proof

In a parallelogram:
$$\angle A=\angle C$$

and
$$\angle B=\angle D$$

But in a cyclic quadrilateral:

$\angle A+\angle C=180^\circ$

Since $\angle A=\angle C$,
$$\angle A+\angle A=180^\circ$$
$$2\angle A=180^\circ$$
$$\angle A=90^\circ$$

Similarly,$$\angle B=90^\circ$$∠B=90∘

Therefore, all the angles of the parallelogram are right angles.

Hence, the parallelogram is a rectangle.

Therefore, a rectangle is the only parallelogram that can be inscribed in a circle.

Hence Proved.

Q15. Show that if a rectangle is inscribed in a circle, then the point of the
intersection of its diagonals must lie at the centre of the circle.

Solution:‘

Proof

In a rectangle:
$$AC=BD$$

and the diagonals bisect each other.

So,
$$AO=OC$$

and$$BO=OD$$

Since diagonals are equal,
$$AC=BD$$

Therefore,$$AO=BO=CO=DO$$

Thus, point $O$O is equidistant from all four vertices $A,B,C,D$.

A point equidistant from all points on a circle is the centre of the circle.

Hence, $O$ is the centre of the circle.

Hence Proved

Q16. Consider all chords of a circle of a fixed length. What is the
shape formed by the midpoints of all these chords?

Solution:

All chords have the same length.

In a circle:

• Equal chords are equidistant from the centre.
• The perpendicular from the centre to a chord passes through its midpoint.

So, the midpoint of every such chord lies at the same fixed distance from the centre.

All the midpoints form another circle having the same centre as the original circle.

This smaller circle is concentric with the original circle.

Ans.

The shape formed by the midpoints is:
$$\boxed{\text{a concentric circle}}$$

Q17. In a circle with centre O, chords AB and AC are congruent. Explain
why this statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.

Solution:

Given

In a circle with centre $O$,$$AB=AC$$

To Prove:

$O$ lies on the angle bisector of $\angle BAC$.

Proof:

Join:$$OB,\ OC,\ OA$$

Since $OB$ and $OC$ are radii of the same circle,
$$OB=OC$$

Given:

$$AB=AC$$

Also,
$$OA=OA$$

(Common side)

Therefore, in triangles $\triangle OBA$ and $\triangle OCA$:
$$OB=OC$$
$$AB=AC$$
$$OA=OA$$

Hence,
$$\triangle OBA \cong \triangle OCA$$

by SSS congruence.

Therefore,
$$\angle BAO=\angle OAC$$

So, line $AO$ bisects $\angle BAC$.

Hence, the centre $O$ lies on the angle bisector of $\angle BAC$.

Hence Proved.

Q18. Two parallel chords of lengths 10 cm and 24 cm are on the same
side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.

Solution:

Given

Two parallel chords are on the same side of the centre.

• Length of first chord $=10\text{ cm}$
• Length of second chord $=24\text{ cm}$
• Distance between the chords $=7\text{ cm}$

Let:

• $r$ be the radius of the circle
• Distance of the longer chord from the centre $=x$
• Distance of the smaller chord from the centre $=x+7$

Since the perpendicular from the centre bisects a chord:

Half of $24\text{ cm}$ chord:
$$12\text{ cm}$$

Half of $10\text{ cm}$ chord:
$$5\text{ cm}$$

Using the Baudhāyana–Pythagoras theorem for the larger chord:

$r^2=x^2+12^2$
$$r^2=x^2+144$$

For the smaller chord:

$r^2=(x+7)^2+5^2$$$r^2=(x+7)^2+25$$

Since both are equal to $r^2$,
$$x^2+144=(x+7)^2+25$$
$$x^2+144=x^2+14x+49+25$$
$$144=14x+74$$
$$70=14x$$
$$x=5$$

Now,
$$r^2=5^2+12^2$$
$$r^2=25+144$$
$$r^2=169$$
$$r=13\text{ cm}$$

Ans.

$$\boxed{13\text{ cm}}$$

Q19. A regular hexagon is inscribed in a circle of radius r. Find the
length of the sides of the hexagon and the distance of each side
from the centre of the circle.

Given

A regular hexagon is inscribed in a circle of radius $r$.

To Find:

1. Length of each side of the hexagon.
2. Distance of each side from the centre of the circle.

Construction

Let $ABCDEF$ be the regular hexagon inscribed in a circle with centre $O$ and radius $r$.

Join $OA$ and $OB$.

Draw $OM \perp AB$, where $M$ is the midpoint of side $AB$.

Solution:

Since $ABCDEF$ is a regular hexagon,$$\angle AOB=\frac{360^\circ}{6}=60^\circ$$

Also,
$$OA=OB=r$$

Thus, in $\triangle AOB$,
$$OA=OB=r \quad \text{and} \quad \angle AOB=60^\circ$$

Hence, $\triangle AOB$ is an equilateral triangle.

Therefore,
$$AB=OA=OB=r$$

So, the length of each side of the hexagon is:
$$\boxed{r}$$

Now, $OM$ is perpendicular to the chord AB from the centre.

Therefore, it bisects the chord $AB$.
$$AM=\frac{AB}{2}=\frac{r}{2}$$

In right-angled $\triangle OAM$,
$$OA=r,\qquad AM=\frac{r}{2}$$

By Pythagoras' theorem,
$$OM^2=OA^2-AM^2$$
$$OM^2=r^2-\left(\frac{r}{2}\right)^2$$
$$OM^2=r^2-\frac{r^2}{4}$$
$$OM^2=\frac{3r^2}{4}$$
$$OM=\frac{\sqrt3\,r}{2}$$

Hence, the distance of each side from the centre is
$$\boxed{\frac{\sqrt3}{2}r}$$

Answer

$$\boxed{\text{Length of each side of the hexagon }=r}$$$$\boxed{\text{Distance of each side from the centre }=\frac{\sqrt3}{2}r}$$

Q20. A quadrilateral MNOP is inscribed in a circle. If MN is a
diameter, what can you say about ∠MOP and ∠MNP? Explain
your reasoning.

Given
A quadrilateral $MNOP$ is inscribed in a circle.
$MN$ is the diameter of the circle.

To Find
The values of:
$$\angle MOP \quad \text{and} \quad \angle MNP$$and explain the reasoning.

Solution:

Since $MN$ is a diameter, any angle subtended by $MN$ at a point on the circle is a right angle.

Therefore, in quadrilateral $MNOP$,
$$\angle MPN = 90^\circ$$(by the theorem: Angle in a semicircle is a right angle)

Now consider the angles $\angle MOP$ and $\angle MNP$.

Both angles subtend the same chord $MP$:

$\angle MOP$ is formed by chords $OM$ and $OP$.

$\angle MNP$ is formed by chords $NM$ and $NP$.

Since angles standing on the same chord of a circle are equal,
$$\angle MOP = \angle MNP$$
Reasoning


Both $\angle MOP$ and $\angle MNP$ intercept the same chord $MP$.

According to the theorem:

Angles in the same segment of a circle are equal.

Hence,
$$\boxed{\angle MOP = \angle MNP}$$
Answer

$$\boxed{\angle MOP = \angle MNP}$$
∠MOP =∠MNP​ because both angles stand on the same chord $MP$ of the circle.

Q21. Let ABCD be a cyclic quadrilateral. Explain why the exterior
angle at any vertex is equal to the interior opposite angle
(e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).

Solution:

Given

$ABCD$ is a cyclic quadrilateral, and $DE$ is the extension of side $CD$.

To Prove

$$\angle ADE=\angle ABC$$

Proof

Since $DE$ is an extension of $CD$,$$\angle ADE+\angle ADC=180^\circ$$

(Linear Pair) ...................................(1)

Since $ABCD$ is a cyclic quadrilateral,$$\angle ABC+\angle ADC=180^\circ$$

...................................(2)

(Opposite angles of a cyclic quadrilateral are supplementary)

From (1) and (2),$$\angle ADE+\angle ADC = \angle ABC+\angle ADC$$

Subtracting $\angle ADC$ from both sides,$$\angle ADE=\angle ABC$$

Hence proved.$$\boxed{\angle ADE=\angle ABC}$$

Q22. “There is no chord of a circle that is longer than its diameter.”
How do you justify this statement?

Solution :