I’m Up and Down, and Round and RoundClass 9 Mathematics NCERT Solutions

Given: Radius of the circle, $r = 13$ cm. Distance of the chord from the centre, $d = 5$ cm.

To Find: The length of the chord.

Formula: The length of a chord ($L$) is given by $L = 2\sqrt{r^2 - d^2}$.

Solution: Let the chord be AB and the center be O. Let M be the midpoint of AB, so $OM = d = 5$ cm. The radius is $OA = r = 13$ cm. In the right-angled triangle $\triangle OMA$: $OA^2 = OM^2 + AM^2$ $13^2 = 5^2 + AM^2$ $169 = 25 + AM^2$ $AM^2 = 169 - 25 = 144$ $AM = \sqrt{144} = 12 \text{ cm}$ The length of the chord is twice the length of AM. $L = 2 \times AM = 2 \times 12 = 24 \text{ cm}$

Final Answer: The length of the chord is 24 cm.

Given: The angle subtended by an arc at the centre is $70^{\circ}$.

To Find: The measure of the angle subtended by the same arc at a point on the remaining part of the circle.

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.

Solution: Let $\theta_c$ be the angle at the centre and $\theta_p$ be the angle at a point on the circumference. According to the theorem, $\theta_c = 2 \times \theta_p$. We are given $\theta_c = 70^{\circ}$. $70^{\circ} = 2 \times \theta_p$ $\theta_p = \frac{70^{\circ}}{2} = 35^{\circ}$

Final Answer: The measure of the angle subtended by the arc at a point on the circle is $35^{\circ}$.

Given: Diameter of the circle = 26 cm. Length of the chord, $L = 24$ cm.

To Find: The distance from the centre of the circle to the chord ($d$).

Solution: First, find the radius of the circle. Radius, $r = \frac{\text{Diameter}}{2} = \frac{26}{2} = 13$ cm.

The perpendicular from the centre to a chord bisects the chord. So, half the length of the chord is $\frac{L}{2} = \frac{24}{2} = 12$ cm.

Let $d$ be the distance from the centre to the chord. We can form a right-angled triangle with the radius as the hypotenuse, the distance $d$ as one leg, and half the chord length as the other leg. Using the Baudhāyana-Pythagoras theorem: $r^2 = d^2 + (\frac{L}{2})^2$ $13^2 = d^2 + 12^2$ $169 = d^2 + 144$ $d^2 = 169 - 144 = 25$ $d = \sqrt{25} = 5 \text{ cm}$

Final Answer: The distance from the centre of the circle to the chord is 5 cm.

Given: Radius of the circle, $r = 15$ cm. Distance from the centre to the chord, $d = 9$ cm.

To Find: The length of the chord ($L$).

Formula: The length of a chord ($L$) is given by $L = 2\sqrt{r^2 - d^2}$.

Solution: Half the length of the chord, let's call it $x$, can be found using the Pythagorean theorem where the radius is the hypotenuse. $r^2 = d^2 + x^2$ $15^2 = 9^2 + x^2$ $225 = 81 + x^2$ $x^2 = 225 - 81 = 144$ $x = \sqrt{144} = 12 \text{ cm}$ The length of the chord is twice this value. $L = 2x = 2 \times 12 = 24 \text{ cm}$

Final Answer: The length of the chord is 24 cm.

To Prove: The perpendicular bisector of a chord passes through the centre of the circle.

Proof: Let AB be a chord of a circle, and let line $l$ be the perpendicular bisector of AB. Let O be the centre of the circle.

By definition, the centre O is a point from which all points on the circumference are equidistant. Since A and B are points on the circle, their distances from the centre O are equal to the radius $r$. So, $OA = OB = r$.

This means that the centre O is a point that is equidistant from the endpoints A and B of the chord.

The locus of all points that are equidistant from two given points (A and B) is the perpendicular bisector of the line segment joining them (AB).

Since the centre O is equidistant from A and B, it must lie on the perpendicular bisector of the segment AB.

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

Given: AB is the diameter of a circle. C is a point on the circumference.

To Find: The measure of $\angle ACB$.

Answer: The measure of $\angle ACB$ is $90^{\circ}$.

Reasoning: The angle subtended by an arc at any point on the circumference is half the angle subtended by the same arc at the centre.

Here, the chord is the diameter AB. The arc corresponding to this chord is a semicircle. The angle subtended by the diameter AB at the centre O is a straight angle, which measures $180^{\circ}$.

Point C is on the circumference. The angle $\angle ACB$ is the angle subtended by the diameter AB at point C.

Using the theorem: $\angle ACB = \frac{1}{2} \times (\text{angle subtended by arc AB at the centre})$ $\angle ACB = \frac{1}{2} \times 180^{\circ}$ $\angle ACB = 90^{\circ}$

This property is often stated as: "The angle in a semicircle is a right angle."

Final Answer: The measure of $\angle ACB$ is $90^{\circ}$.

Given: ABCD is a cyclic quadrilateral. $\angle A = 75^{\circ}$ $\angle B = 110^{\circ}$

To Find: The measures of $\angle C$ and $\angle D$.

Theorem: The sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$.

Solution: For a cyclic quadrilateral ABCD, the pairs of opposite angles are ($\angle A, \angle C$) and ($\angle B, \angle D$).

Final Answer: The measure of $\angle C$ is $105^{\circ}$, and the measure of $\angle D$ is $70^{\circ}$.

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

To Find: The value of $x$, and the measures of $\angle P$ and $\angle R$.

Theorem: The sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$.

Solution: Angles P and R are opposite angles in the cyclic quadrilateral PQRS. Therefore, $\angle P + \angle R = 180^{\circ}$.

Substitute the given expressions for the angles: $(2x + 10) + (3x - 20) = 180$ Combine like terms: $5x - 10 = 180$ Add 10 to both sides: $5x = 190$ Divide by 5: $x = \frac{190}{5} = 38$

Now, find the measure of each angle by substituting $x=38$.

Measure of $\angle P$: $\angle P = 2x + 10 = 2(38) + 10 = 76 + 10 = 86^{\circ}$

Measure of $\angle R$: $\angle R = 3x - 20 = 3(38) - 20 = 114 - 20 = 94^{\circ}$

Verification: $\angle P + \angle R = 86^{\circ} + 94^{\circ} = 180^{\circ}$. The result is correct.

Final Answer: The value of $x$ is 38. The measures of the angles are $\angle P = 86^{\circ}$ and $\angle R = 94^{\circ}$.

Given: Length of the chord, $L = 16$ cm. Distance of the chord from the centre, $d = 6$ cm.

To Find: The radius of the circle ($r$).

Solution: The perpendicular from the centre to a chord bisects the chord. Half the length of the chord is $\frac{L}{2} = \frac{16}{2} = 8$ cm.

We can form a right-angled triangle with the radius as the hypotenuse, the distance $d$ as one leg, and half the chord length as the other leg. Using the Baudhāyana-Pythagoras theorem: $r^2 = d^2 + (\frac{L}{2})^2$ $r^2 = 6^2 + 8^2$ $r^2 = 36 + 64$ $r^2 = 100$ $r = \sqrt{100} = 10 \text{ cm}$

Final Answer: The radius of the circle is 10 cm.

Given: A cyclic quadrilateral with side lengths $a=5, b=5, c=12, d=12$.

To Find: The area of the quadrilateral.

Formula: The area of a cyclic quadrilateral can be found using Brahmagupta's formula: $Area = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semi-perimeter, $s = \frac{a+b+c+d}{2}$.

Solution: First, calculate the semi-perimeter $s$: $s = \frac{5 + 5 + 12 + 12}{2} = \frac{34}{2} = 17 \text{ units}$

Now, apply Brahmagupta's formula: $Area = \sqrt{(17-5)(17-5)(17-12)(17-12)}$ $Area = \sqrt{12 \times 12 \times 5 \times 5}$ $Area = \sqrt{12^2 \times 5^2}$ $Area = \sqrt{(12 \times 5)^2}$ $Area = 12 \times 5 = 60 \text{ square units}$

Final Answer: The area of the cyclic quadrilateral is 60 square units.

To determine: If the circumcentre of a cyclic quadrilateral lies inside or outside it.

Method: The position of the circumcentre of a cyclic quadrilateral depends on whether the quadrilateral contains an obtuse angle when considered as a vertex of a triangle formed by a diagonal.

A simple and effective way is to analyze the triangles formed by the diagonals of the quadrilateral.

Let the cyclic quadrilateral be ABCD. Let the side lengths be $a=AB, b=BC, c=CD, d=DA$. We would also need the lengths of the diagonals $p=AC$ and $q=BD$.

Summary of the method: To determine if the circumcentre is inside or outside, one must know the side lengths and at least one diagonal's length. Then, use the Law of Cosines to check if any of the quadrilateral's four angles is obtuse. If any angle is obtuse, the circumcentre is outside. If all angles are acute or right, the circumcentre is inside or on the boundary.

Given: Two chords AB and CD of a circle are of equal length, so $AB = CD$. They intersect at a point P inside the circle. Chord AB is divided into segments AP and PB. Chord CD is divided into segments CP and PD.

To Prove: The corresponding segments are equal. That is, $AP = CP$ and $PB = PD$, or $AP = PD$ and $PB = CP$.

This shows that the corresponding segments of the chords are equal.

To Construct: A circle with a chord of 6 cm at a distance of 3 cm from the centre.

Analysis: First, we need to determine the radius of this circle. Let the chord be AB, with length 6 cm. Let the centre be O, and the distance from O to AB be OM = 3 cm. The point M will be the midpoint of AB, so $AM = \frac{6}{2} = 3$ cm. The radius of the circle is OA. In the right-angled triangle $\triangle OMA$: $r^2 = OA^2 = OM^2 + AM^2$ $r^2 = 3^2 + 3^2 = 9 + 9 = 18$ $r = \sqrt{18} = 3\sqrt{2} \approx 4.24 \text{ cm}$

Result: The constructed circle has a chord AB of length 6 cm, which is at a perpendicular distance of 3 cm from the centre O.

To Prove: If a parallelogram is inscribed in a circle, it must be a rectangle.

Proof: Let ABCD be a parallelogram inscribed in a circle. This means ABCD is a cyclic parallelogram.

Properties of a parallelogram:

Properties of a cyclic quadrilateral:

Let's combine these properties. From the parallelogram property, we have $\angle A = \angle C$. From the cyclic quadrilateral property, we have $\angle A + \angle C = 180^{\circ}$.

Substitute $\angle A$ for $\angle C$ in the second equation: $\angle A + \angle A = 180^{\circ}$ $2\angle A = 180^{\circ}$ $\angle A = 90^{\circ}$ Since $\angle A = \angle C$, it follows that $\angle C = 90^{\circ}$.

Similarly, for the other pair of opposite angles: From the parallelogram property, $\angle B = \angle D$. From the cyclic quadrilateral property, $\angle B + \angle D = 180^{\circ}$.

Substitute $\angle B$ for $\angle D$: $\angle B + \angle B = 180^{\circ}$ $2\angle B = 180^{\circ}$ $\angle B = 90^{\circ}$ Since $\angle B = \angle D$, it follows that $\angle D = 90^{\circ}$.

Thus, all four angles of the parallelogram ABCD are $90^{\circ}$. A parallelogram with all four angles being right angles is, by definition, a rectangle.

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

Given: A rectangle ABCD is inscribed in a circle.

To Prove: The point of intersection of the diagonals of the rectangle is the centre of the circle.

Proof: Let the rectangle be ABCD. By definition, all its angles are right angles: $\angle A = \angle B = \angle C = \angle D = 90^{\circ}$. The vertices A, B, C, and D lie on the circle.

Consider the diagonal AC of the rectangle. This diagonal is a chord of the circle. The angle subtended by this chord at point B on the circumference is $\angle ABC$. We know $\angle ABC = 90^{\circ}$.

A property of circles states that the angle in a semicircle is a right angle. The converse is also true: if a chord subtends a right angle at any point on the circumference, the chord must be a diameter of the circle. Since the chord AC subtends a $90^{\circ}$ angle at point B, AC must be a diameter of the circle.

Similarly, consider the diagonal BD of the rectangle. This is also a chord of the circle. The angle subtended by this chord at point A on the circumference is $\angle DAB$. We know $\angle DAB = 90^{\circ}$. Since the chord BD subtends a $90^{\circ}$ angle at point A, BD must also be a diameter of the circle.

So, both diagonals AC and BD are diameters of the same circle. The centre of a circle is the point of intersection of all its diameters. Therefore, the point where the diagonals AC and BD intersect is the centre of the circle.

Given: A circle with centre O and radius $r$. A set of chords, all having the same fixed length, $L$.

To Find: The shape (locus) formed by the midpoints of all these chords.

Solution: Let AB be any chord of length $L$. Let M be its midpoint. The distance of this chord from the centre is the length of the perpendicular segment OM.

In the right-angled triangle $\triangle OMA$, by the Baudhāyana-Pythagoras theorem: $OA^2 = OM^2 + AM^2$ Here, $OA = r$ (radius) and $AM = L/2$ (since M is the midpoint of the chord of length L). $r^2 = OM^2 + (L/2)^2$ Solving for OM: $OM^2 = r^2 - (L/2)^2$ $OM = \sqrt{r^2 - (L/2)^2}$

Since the radius $r$ and the chord length $L$ are fixed constants, the value of OM is also a constant. Let's call this constant distance $d$. $d = \sqrt{r^2 - (L/2)^2}$

This means that the midpoint M of any chord of length $L$ is always at a fixed distance $d$ from the centre O.

The locus of all points that are at a fixed distance from a fixed point is the definition of a circle.

Therefore, the shape formed by the midpoints of all chords of a fixed length is a circle. This circle is concentric with the original circle and has a radius of $d = \sqrt{r^2 - (L/2)^2}$.

Final Answer: The shape formed by the midpoints of all these chords is a circle.

Given: A circle with centre O. Two chords AB and AC starting from the same point A. The chords are congruent, meaning $AB = AC$.

To Explain: Why the centre O lies on the angle bisector of $\angle BAC$.

Explanation: To show that O lies on the angle bisector of $\angle BAC$, we need to prove that the line segment OA bisects $\angle BAC$, which means we need to show that $\angle OAB = \angle OAC$.

Let's consider the two triangles formed by joining the centre O to the vertices A, B, and C: $\triangle OAB$ and $\triangle OAC$.

We will compare these two triangles:

By the Side-Side-Side (SSS) congruence criterion, the three sides of $\triangle OAB$ are equal to the three corresponding sides of $\triangle OAC$. Therefore, $\triangle OAB \cong \triangle OAC$.

Since the two triangles are congruent, their corresponding parts must be equal. The angles corresponding to the equal sides OB and OC are $\angle OAB$ and $\angle OAC$, respectively. Therefore, $\angle OAB = \angle OAC$.

Since $\angle OAB = \angle OAC$, the line segment OA divides $\angle BAC$ into two equal angles. This means OA is the angle bisector of $\angle BAC$.

Because the centre O is a point on the line segment OA, the centre of the circle lies on the angle bisector of $\angle BAC$.

Given: Length of first chord, $L_1 = 24$ cm. Length of second chord, $L_2 = 10$ cm. The chords are parallel and on the same side of the centre. Distance between the chords is 7 cm.

To Find: The radius of the circle ($r$).

Solution: Let the centre of the circle be O. Let the chords be AB (length 24 cm) and CD (length 10 cm). Let the perpendicular from O to the chords meet AB at M and CD at N. Since the chords are parallel, O, N, M are collinear. The longer chord is closer to the centre. So, O is on one side, then N, then M.

Distance from centre to chord AB is OM. Distance from centre to chord CD is ON. We are given that the distance between the chords is $MN = 7$ cm. So, $ON = OM + 7$.

Half-length of chord AB is $AM = 24/2 = 12$ cm. Half-length of chord CD is $CN = 10/2 = 5$ cm.

Let the distance $OM = x$. Then $ON = x + 7$.

Consider the right-angled triangle $\triangle OMA$: $r^2 = OM^2 + AM^2$ $r^2 = x^2 + 12^2 = x^2 + 144 \quad \dots(1)$

Consider the right-angled triangle $\triangle ONC$: $r^2 = ON^2 + CN^2$ $r^2 = (x+7)^2 + 5^2 = (x^2 + 14x + 49) + 25 = x^2 + 14x + 74 \quad \dots(2)$

Since both expressions equal $r^2$, we can equate them: $x^2 + 144 = x^2 + 14x + 74$ Subtract $x^2$ from both sides: $144 = 14x + 74$ $14x = 144 - 74$ $14x = 70$ $x = \frac{70}{14} = 5$ So, the distance of the longer chord from the centre is $OM = 5$ cm.

Now, substitute the value of $x$ back into equation (1) to find the radius $r$: $r^2 = x^2 + 144$ $r^2 = 5^2 + 144$ $r^2 = 25 + 144 = 169$ $r = \sqrt{169} = 13 \text{ cm}$

Final Answer: The radius of the circle is 13 cm.

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

1. Length of the sides: A regular hexagon can be divided into six congruent equilateral triangles, with one common vertex at the centre of the circle (O) and the bases of the triangles forming the sides of the hexagon.

Let one side of the hexagon be AB. Consider the triangle $\triangle OAB$. The sides OA and OB are radii of the circle, so $OA = OB = r$. The angle subtended by each side of a regular hexagon at the centre is $\frac{360^{\circ}}{6} = 60^{\circ}$. So, $\angle AOB = 60^{\circ}$.

Since $\triangle OAB$ is an isosceles triangle with a vertex angle of $60^{\circ}$, the other two angles are equal: $\angle OAB = \angle OBA = \frac{180^{\circ} - 60^{\circ}}{2} = 60^{\circ}$. As all three angles are $60^{\circ}$, $\triangle OAB$ is an equilateral triangle. Therefore, all its sides are equal: $AB = OA = OB = r$.

The length of a side of the hexagon is equal to the radius of the circle.

2. Distance of each side from the centre: The distance of a side from the centre is the length of the perpendicular from the centre to that side. Let M be the midpoint of side AB. The distance is OM. OM is the altitude of the equilateral triangle $\triangle OAB$.

In the right-angled triangle $\triangle OMA$:

By the Baudhāyana-Pythagoras theorem: $OA^2 = OM^2 + AM^2$ $r^2 = OM^2 + (\frac{r}{2})^2$ $r^2 = OM^2 + \frac{r^2}{4}$ $OM^2 = r^2 - \frac{r^2}{4} = \frac{3r^2}{4}$ $OM = \sqrt{\frac{3r^2}{4}} = \frac{\sqrt{3}r}{2}$

Final Answer: The length of the sides of the hexagon is $r$. The distance of each side from the centre of the circle is $\frac{\sqrt{3}}{2}r$.

Given: A cyclic quadrilateral MNOP. MN is a diameter of the circle.

To Find: Information about $\angle MOP$ and $\angle MNP$.

Reasoning: The property that the angle subtended by a diameter at any point on the circumference is a right angle ($90^{\circ}$) is key here.

1. About $\angle MOP$: This is not a valid angle in the standard sense of the theorem. The angle subtended by the diameter MN would be at vertex O or vertex P. Let's assume the question meant $\angle MON$ and $\angle MPN$.

2. About $\angle MPN$: The angle $\angle MPN$ is subtended by the diameter MN at the vertex P, which lies on the circumference. According to the theorem, the angle in a semicircle is a right angle. Therefore, $\angle MPN = 90^{\circ}$.

3. About $\angle MON$: The angle $\angle MON$ is subtended by the diameter MN at the vertex O, which lies on the circumference. According to the same theorem, the angle in a semicircle is a right angle. Therefore, $\angle MON = 90^{\circ}$.

Conclusion: Both vertices P and O lie on the circumference of the circle. The angles at these vertices that are subtended by the diameter MN are both right angles.

Final Answer: Both $\angle MPN$ and $\angle MON$ are right angles. The reason is that they are angles subtended by a diameter (MN) at points (P and O) on the circumference of the circle. The angle in a semicircle is always $90^{\circ}$.

Given: ABCD is a cyclic quadrilateral. Side DC is extended to a point E, creating an exterior angle $\angle BCE$. (Let's assume the vertex is C and the extended side is DC to E for clarity).

To Explain: Why the exterior angle ($\angle BCE$) is equal to the interior opposite angle ($\angle BAD$).

This shows that the exterior angle at vertex C is equal to the interior opposite angle at vertex A. The same logic can be applied to any vertex of the cyclic quadrilateral.

To Justify: No chord of a circle is longer than its diameter.

Justification: There are a few ways to justify this statement.

Method 1: Using the chord length formula Let a circle have radius $r$. The diameter has length $D = 2r$. Let any chord have length $L$. Let the perpendicular distance of this chord from the centre be $d$.

The relationship between these quantities is given by the Baudhāyana-Pythagoras theorem: $r^2 = d^2 + (L/2)^2$ Solving for the chord length $L$: $(L/2)^2 = r^2 - d^2$ $L/2 = \sqrt{r^2 - d^2}$ $L = 2\sqrt{r^2 - d^2}$

The distance $d$ is a real-world distance, so $d \ge 0$, which means $d^2 \ge 0$. Therefore, $r^2 - d^2 \le r^2$. Taking the square root, $\sqrt{r^2 - d^2} \le r$. Multiplying by 2, $2\sqrt{r^2 - d^2} \le 2r$. This means $L \le 2r$, or $L \le D$.

The length of any chord is less than or equal to the diameter. The equality $L=D$ occurs only when $d=0$, which is the case for a chord that passes through the centre, i.e., a diameter itself.

Method 2: Using the Triangle Inequality Let AB be any chord of the circle with centre O. Consider the triangle $\triangle OAB$. The sides are OA, OB, and AB. $OA = r$ and $OB = r$. By the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side. $OA + OB \ge AB$ $r + r \ge AB$ $2r \ge AB$ Since the diameter $D = 2r$, this gives $D \ge AB$. This shows that the length of any chord AB is less than or equal to the diameter. The equality holds only if the 'triangle' is degenerate, meaning O lies on the line segment AB. This is precisely the case when AB is a diameter.

Conclusion: Both methods show that a chord's length can never exceed the diameter's length.

Given: A circle with centre O. A point A inside the circle.

To Prove: The shortest chord passing through A is the one perpendicular to the line segment OA.

Proof: Let PQ be any chord that passes through the point A. Let M be the midpoint of the chord PQ. The line OM is perpendicular to the chord PQ, so $\triangle OMA$ is a right-angled triangle with the right angle at M.

The length of the chord PQ is given by the formula $L = 2\sqrt{r^2 - d^2}$, where $r$ is the radius and $d = OM$ is the distance of the chord from the centre. To find the shortest chord, we need to minimize the length $L$. This is equivalent to maximizing the distance $d = OM$.

Now, consider the right-angled triangle $\triangle OMA$. The hypotenuse is OA (the distance from the centre to the fixed point A). The legs are OM and AM. By the Pythagorean theorem, $OA^2 = OM^2 + AM^2$. Since $AM^2 \ge 0$, we have $OA^2 \ge OM^2$, which implies $OA \ge OM$.

This shows that the distance of any chord passing through A from the centre (OM) is always less than or equal to the distance OA. The maximum possible value for the distance $OM$ is $OA$. This maximum occurs when the leg AM is zero, which means the point M (the midpoint of the chord) coincides with the point A.

If the midpoint M of the chord PQ is the point A, then the line OM becomes the line OA. Since OM is always perpendicular to the chord PQ, it follows that when M coincides with A, the line OA is perpendicular to the chord PQ.

So, the chord is shortest when its distance from the centre is maximized, which happens when the chord is perpendicular to the line segment OA.

To Justify: The angle in a semicircle is $90^{\circ}$.

Figure Description: The figure shows a circle with centre O and diameter AB. C is a point on the circumference. A line segment is drawn from C to the centre O, dividing the triangle ABC into two smaller triangles, $\triangle AOC$ and $\triangle BOC$.

This justifies that the angle ($\angle ACB$) in a semicircle (the arc on which C lies, with diameter AB) is a right angle.

Given: A circle with diameter AB. Chords $CC'$ and $DD'$ are both perpendicular to the diameter AB. M is the midpoint of chord CD. M' is the midpoint of chord C'D'.

To Prove: The segment MM' is perpendicular to AB.

Proof using Coordinate Geometry: Let the centre of the circle be the origin O(0, 0) and the radius be $r$. Let the diameter AB lie along the x-axis. So, $A=(-r, 0)$ and $B=(r, 0)$.

Since chord $CC'$ is perpendicular to AB (the x-axis), it is a vertical line. Let its equation be $x = p$. The points C and C' will have coordinates $C(p, y_c)$ and $C'(p, -y_c)$, where $y_c = \sqrt{r^2 - p^2}$. The diameter AB is the axis of symmetry for the chord $CC'$.

Similarly, since chord $DD'$ is perpendicular to AB, it is also a vertical line. Let its equation be $x = q$. The points D and D' will have coordinates $D(q, y_d)$ and $D'(q, -y_d)$, where $y_d = \sqrt{r^2 - q^2}$.

Now, we find the coordinates of the midpoints M and M'.

Midpoint M of chord CD: The coordinates of C are $(p, \sqrt{r^2-p^2})$. The coordinates of D are $(q, \sqrt{r^2-q^2})$. Using the midpoint formula, $M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$: $M = \left( \frac{p+q}{2}, \frac{\sqrt{r^2-p^2} + \sqrt{r^2-q^2}}{2} \right)$

Midpoint M' of chord C'D': The coordinates of C' are $(p, -\sqrt{r^2-p^2})$. The coordinates of D' are $(q, -\sqrt{r^2-q^2})$. Using the midpoint formula: $M' = \left( \frac{p+q}{2}, \frac{-\sqrt{r^2-p^2} - \sqrt{r^2-q^2}}{2} \right)$

Analyze the segment MM': The coordinates of M are $(x_M, y_M) = \left( \frac{p+q}{2}, \frac{\sqrt{r^2-p^2} + \sqrt{r^2-q^2}}{2} \right)$. The coordinates of M' are $(x_{M'}, y_{M'}) = \left( \frac{p+q}{2}, -\frac{\sqrt{r^2-p^2} + \sqrt{r^2-q^2}}{2} \right)$.

We observe that the x-coordinate of M is the same as the x-coordinate of M'. $x_M = x_{M'} = \frac{p+q}{2}$

A line segment joining two points with the same x-coordinate is a vertical line. The equation of the line passing through M and M' is $x = \frac{p+q}{2}$.

The diameter AB lies on the x-axis (equation $y=0$). A vertical line is always perpendicular to a horizontal line.

Therefore, the segment MM' is perpendicular to the diameter AB.

To Justify: The sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$.

Figure Description: The figure shows a cyclic quadrilateral ABCD with its centre O. Lines are drawn from the centre O to all four vertices A, B, C, and D. This divides the quadrilateral into four isosceles triangles: $\triangle OAB, \triangle OBC, \triangle OCD, \triangle ODA$.

This justification shows that the sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$. (This method assumes the centre O is inside the quadrilateral).

Given: In $\triangle ABC$: $\mathrm{AB}=5 \mathrm{~cm}$ $\angle \mathrm{A}=70^{\circ}$ $\angle \mathrm{B}=60^{\circ}$

Solution: First, we find the third angle of the triangle: $\angle C = 180^{\circ} - (\angle A + \angle B)$ $\angle C = 180^{\circ} - (70^{\circ} + 60^{\circ})$ $\angle C = 180^{\circ} - 130^{\circ} = 50^{\circ}$ Since all angles ($70^{\circ}, 60^{\circ}, 50^{\circ}$) are less than $90^{\circ}$, $\triangle ABC$ is an acute-angled triangle.

Conclusion: For any acute-angled triangle, the circumcentre always lies inside the triangle. Since $\triangle ABC$ is an acute-angled triangle, its circumcentre lies inside the triangle.

Final Answer: The centre of the circumcircle is inside the triangle.

Given: In $\triangle ABC$: $\mathrm{AB}=5 \mathrm{~cm}$ $\angle \mathrm{A}=100^{\circ}$ $\mathrm{AC}=4 \mathrm{~cm}$

Solution: The triangle has an angle $\angle A = 100^{\circ}$, which is greater than $90^{\circ}$. Therefore, $\triangle ABC$ is an obtuse-angled triangle.

Conclusion: For any obtuse-angled triangle, the circumcentre always lies outside the triangle. Since $\triangle ABC$ is an obtuse-angled triangle, its circumcentre lies outside the triangle.

Final Answer: The centre of the circumcircle is outside the triangle.