Chapter 10 Circles

A circle can have infinitely many tangents.

(i) A tangent to a circle intersects it in one point (s).

(ii) A line intersecting a circle in two points is called a secant.

(iii) A circle can have two parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called point of contact.

Given:

A circle with centre O and radius OP = 5 cm.

PQ is a tangent to the circle at point P.

A line from the centre O meets the tangent at Q, such that OQ = 12 cm.

To Find:

The length of the tangent segment PQ.

Solution:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, the radius OP is perpendicular to the tangent PQ at P.

Consider the triangle $\triangle OPQ$. Since $\angle OPQ = 90^\circ$, it is a right-angled triangle.

By the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $\triangle OPQ$, OQ is the hypotenuse.

Substitute the given values, OP = 5 cm and OQ = 12 cm:

Subtract 25 from both sides to find $PQ^2$:

Take the square root of both sides to find the length PQ:

The length of PQ is $\sqrt{119}$ cm.

Conclusion:

Comparing the result with the given options, the correct option is (D).

Steps for Construction:

1. Draw a circle with any centre O and a suitable radius.

2. Draw a straight line $l$ outside, on, or intersecting the circle. This is the given line.

3. To draw a tangent parallel to line $l$: Draw a line perpendicular to $l$ passing through the centre O. This line will intersect the circle at two points. Draw a line through one of these intersection points that is perpendicular to the radius at that point. This new line will be a tangent to the circle and parallel to $l$. Let's call this line $m$.

4. To draw a secant parallel to line $l$: Draw another line perpendicular to $l$ passing through the interior of the circle. This line will intersect the circle at two points. Draw a line through these two intersection points. This new line will be a secant to the circle and parallel to $l$. Let's call this line $n$.

Description of the Figure:

The figure shows a circle with centre O.

A line $l$ is drawn somewhere in the plane.

A line $m$ is drawn such that $m || l$ and $m$ touches the circle at exactly one point (is a tangent).

A line $n$ is drawn such that $n || l$ and $n$ intersects the circle at two distinct points (is a secant).

There is a visual representation showing the circle, the original line, and the two lines parallel to it, one touching the circle and the other passing through it.

Given:

Two concentric circles with common centre O.

A chord AB of the larger circle touches the smaller circle at point P.

To Prove:

The chord AB is bisected at the point of contact P, which means AP = PB.

Construction:

Join the centre O to the point of contact P. Join OA and OB (optional, but can help visualize the larger circle properties).

Proof:

We are given that AB is a tangent to the smaller circle at point P.

OP is the radius of the smaller circle to the point of contact P.

According to the theorem that states the tangent at any point of a circle is perpendicular to the radius through the point of contact, we have:

This implies that $\angle OPA = 90^\circ$ and $\angle OPB = 90^\circ$.

Now, consider the larger circle.

AB is a chord of the larger circle.

OP is a line segment from the centre O of the larger circle to the chord AB.

We have shown that $OP \perp AB$.

According to the theorem that states the perpendicular from the centre of a circle to a chord bisects the chord, the point P must be the midpoint of the chord AB.

Therefore, P bisects the chord AB.

This means that the length of segment AP is equal to the length of segment PB.

Hence, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

This completes the proof.

Given:

A circle with centre O.

TP and TQ are two tangents drawn from an external point T to the circle.

P and Q are the points of contact of the tangents.

To Prove:

$\angle PTQ = 2 \angle OPQ$

Proof:

We know that the tangents drawn from an external point to a circle are equal in length.

So, in the figure:

Consider the triangle $\triangle TPQ$. Since $TP = TQ$, $\triangle TPQ$ is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal.

Therefore,

Let $\angle PTQ = \theta$.

The sum of angles in a triangle is $180^\circ$. In $\triangle TPQ$:

Substitute $\angle PTQ = \theta$ and $\angle TPQ = \angle TQP$:

We also know that the radius through the point of contact is perpendicular to the tangent.

So, the radius OP is perpendicular to the tangent TP at P.

From the figure, $\angle OPT$ is the sum of $\angle OPQ$ and $\angle TPQ$:

Substitute the value of $\angle OPT$ and the expression for $\angle TPQ$ from (i):

Now, solve for $\angle OPQ$:

Since we defined $\theta = \angle PTQ$, we can substitute this back into equation (ii):

Multiply both sides by 2:

Rearranging the terms, we get the desired result:

Hence, proved.

Given:

A circle with centre O.

Chord PQ of length $8$ cm.

Radius of the circle $OP = OQ = 5$ cm.

Tangents at P and Q intersect at point T.

To Find:

The length of the tangent TP.

Solution:

We know that the tangents drawn from an external point to a circle are equal in length. Therefore, $TP = TQ$.

The line segment joining the centre to the external point T bisects the angle between the tangents ($\angle PTQ$) and also bisects the chord PQ perpendicularly.

So, the line segment TO bisects the chord PQ at point R, and $TO \perp PQ$.

This means R is the midpoint of PQ, and $\angle PRO = \angle TRO = 90^\circ$.

Length of PQ = $8$ cm.

Consider the right-angled triangle $\triangle ORP$. We have $OP = 5$ cm (radius) and $PR = 4$ cm.

By the Pythagorean theorem in $\triangle ORP$:

Now consider the triangles $\triangle RPT$ and $\triangle OPT$.

$\angle TRP = 90^\circ$ (Since $TO \perp PQ$)

$\angle OPT = 90^\circ$ (Radius is perpendicular to tangent at point of contact)

$\angle RTP = \angle OTP$ (Common angle to both triangles)

Therefore, by the AA similarity criterion, we have:

Since the triangles are similar, the ratio of their corresponding sides is equal:

From the ratio $\frac{RT}{PT} = \frac{RP}{OP}$:

This gives us $TR = \frac{4}{5} TP$.

From the ratio $\frac{PT}{OT} = \frac{RP}{OP}$:

This gives us $OT = \frac{5}{4} TP$.

From the figure, point R lies on the line segment OT. Therefore, $OT = OR + TR$.

Substitute the values we found for OT, OR, and TR:

Now, solve for TP:

To subtract the terms with TP, find a common denominator, which is 20:

Multiply both sides by $\frac{20}{9}$ to isolate TP:

Simplify the expression by cancelling common factors:

Final Answer:

The length of the tangent TP is $\frac{20}{3}$ cm.

Given:

Length of the tangent from point Q to the circle, PQ = 24 cm.

Distance of point Q from the centre O, OQ = 25 cm.

To Find:

The radius of the circle, OP.

Justification and Solution:

Let O be the centre of the circle and P be the point of contact of the tangent from Q to the circle.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, the radius OP is perpendicular to the tangent PQ at P.

This means that the triangle $\triangle OPQ$ is a right-angled triangle with the right angle at P ($\angle OPQ = 90^\circ$).

We can use the Pythagorean theorem in the right-angled triangle $\triangle OPQ$. The hypotenuse is OQ (the side opposite the right angle).

Substitute the given values, $PQ = 24$ and $OQ = 25$:

Subtract 576 from both sides to find $OP^2$:

Take the square root of both sides to find OP:

The radius of the circle is 7 cm.

Final Answer:

The correct option is (A) 7 cm.

Given:

TP and TQ are tangents to a circle with centre O.

$\angle POQ = 110^\circ$.

To Find:

The measure of $\angle PTQ$.

Justification and Solution:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, the radius OP is perpendicular to the tangent TP at P, and the radius OQ is perpendicular to the tangent TQ at Q.

Consider the quadrilateral OPTQ. The sum of the interior angles in a quadrilateral is $360^\circ$.

Substitute the known values: $\angle OPT = 90^\circ$, $\angle TQO = 90^\circ$, and $\angle QOP = 110^\circ$:

Combine the constant terms:

Subtract $290^\circ$ from both sides to solve for $\angle PTQ$:

The measure of $\angle PTQ$ is $70^\circ$.

Final Answer:

The correct option is (B) $70^\circ$.

Given:

PA and PB are tangents from external point P to a circle with centre O.

The angle between the tangents, $\angle APB = 80^\circ$.

To Find:

The measure of $\angle POA$.

Justification and Solution:

We know that the tangents drawn from an external point to a circle are equal in length.

We also know that the line segment joining the centre to the external point bisects the angle between the tangents.

Thus, the line segment OP bisects $\angle APB$ and also bisects the angle subtended by the chord of contact at the centre ($\angle AOB$).

Since $\angle APB = 80^\circ$, we have:

We know that the radius through the point of contact is perpendicular to the tangent.

So, the radius OA is perpendicular to the tangent PA at A.

Now, consider the right-angled triangle $\triangle OAP$.

The sum of angles in a triangle is $180^\circ$.

Substitute the values $\angle OAP = 90^\circ$ and $\angle APO = 40^\circ$ (from (i)):

Combine the constant terms:

Subtract $130^\circ$ from both sides to solve for $\angle POA$:

Alternatively:

Consider $\triangle OAP$ and $\triangle OBP$.

Therefore, by SSS congruence rule:

By CPCT (Corresponding Parts of Congruent Triangles):

Also, $\angle OAP = \angle OBP = 90^\circ$ (Radius $\perp$ Tangent).

In quadrilateral OAPB, the sum of interior angles is $360^\circ$.

Substitute the known values:

Since $\angle AOB = \angle POA + \angle POB$ and $\angle POA = \angle POB$, we have $\angle AOB = 2 \angle POA$.

Final Answer:

The measure of $\angle POA$ is $50^\circ$.

The correct option is (A) $50^\circ$.

Given:

A circle with centre O.

AB is a diameter of the circle.

Let line $l$ be the tangent to the circle at the end point A of the diameter.

Let line $m$ be the tangent to the circle at the other end point B of the diameter.

To Prove:

The tangents $l$ and $m$ are parallel, i.e., $l || m$.

Proof:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

The radius OA is drawn to the point of contact A of the tangent $l$.

Therefore, the radius OA is perpendicular to the tangent line $l$ at A.

Since A, O, and B are collinear (as AB is a diameter), the diameter AB is perpendicular to the tangent $l$ at A.

Let's consider the angle formed by the tangent $l$ and the line AB. Let X be a point on line $l$ such that X and B are on opposite sides of A.

Similarly, the radius OB is drawn to the point of contact B of the tangent $m$.

Therefore, the radius OB is perpendicular to the tangent line $m$ at B.

Since A, O, and B are collinear, the diameter AB is perpendicular to the tangent $m$ at B.

Let Y be a point on line $m$ such that Y and A are on opposite sides of B.

Now, consider the lines $l$ and $m$ intersected by the transversal line AB.

The angles $\angle BAX$ (from equation (i)) and $\angle ABY$ (from equation (ii)) are alternate interior angles formed by the transversal AB with lines $l$ and $m$.

From equations (i) and (ii), we have $\angle BAX = 90^\circ$ and $\angle ABY = 90^\circ$.

Since the alternate interior angles formed by the transversal AB are equal, the lines $l$ and $m$ must be parallel.

Therefore, the tangents drawn at the ends of a diameter of a circle are parallel.

This completes the proof.

Given:

A circle with centre O.

A tangent line $l$ touches the circle at point P.

To Prove:

The line perpendicular to the tangent $l$ at the point of contact P passes through the centre O.

Proof:

Let the tangent to the circle at point P be the line $l$.

OP is the radius of the circle joining the centre O to the point of contact P.

We know from a theorem that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, the radius OP is perpendicular to the tangent line $l$ at the point of contact P.

Now, consider the line OP. This line passes through the centre O and is perpendicular to the tangent $l$ at P.

From basic geometry, we know that through a given point on a given line, there exists one and only one line perpendicular to the given line.

In our case, the given line is the tangent $l$, and the given point on the line is the point of contact P.

The line OP is one line that is perpendicular to $l$ at P.

According to the uniqueness principle mentioned above, any other line that is perpendicular to the tangent $l$ at the point P must coincide with the line OP.

Since the line OP passes through the centre O (by its construction as a radius connecting the centre to a point on the circle), it logically follows that the perpendicular line to the tangent at the point of contact P must pass through the centre O.

This completes the proof.

Given:

A point A is at a distance of 5 cm from the centre O of the circle. OA = 5 cm.

Let P be the point of contact of the tangent from A to the circle.

The length of the tangent from A to the circle, AP = 4 cm.

To Find:

The radius of the circle, OP.

Solution:

Let O be the centre of the circle and P be the point of contact of the tangent from point A to the circle.

We are given OA = 5 cm and AP = 4 cm. OP is the radius we need to find.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, the radius OP is perpendicular to the tangent AP at the point of contact P.

This means that $\triangle OPA$ is a right-angled triangle with the right angle at P ($\angle OPA = 90^\circ$).

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $\triangle OPA$, the hypotenuse is OA (the side opposite the right angle).

So, by the Pythagorean theorem:

Substitute the given values, $AP = 4$ and $OA = 5$:

Calculate the squares:

Subtract 16 from both sides to solve for $OP^2$:

Take the square root of both sides to find the radius OP:

The radius of the circle is 3 cm.

Final Answer:

The radius of the circle is 3 cm.

Given:

Two concentric circles with common centre O.

Radius of the smaller circle, $r = 3$ cm.

Radius of the larger circle, $R = 5$ cm.

A chord AB of the larger circle touches the smaller circle at point P.

To Find:

The length of the chord AB.

Solution:

Let O be the centre of the two concentric circles.

The radius of the smaller circle is $OP = r = 3$ cm, where P is the point of contact of the chord AB with the smaller circle.

The radius of the larger circle is $OA = OB = R = 5$ cm.

We know that the tangent to a circle is perpendicular to the radius through the point of contact.

Since AB is a tangent to the smaller circle at P, the radius OP is perpendicular to the chord AB.

This means that the angle $\angle OPA = 90^\circ$.

Consider the triangle $\triangle OPA$. It is a right-angled triangle at P.

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $\triangle OPA$, OA is the hypotenuse.

Substitute the known values, $OP = 3$ and $OA = 5$:

Calculate the squares:

Subtract 9 from both sides to solve for $AP^2$:

Take the square root of both sides to find the length AP:

Now, consider the larger circle. AB is a chord of the larger circle, and OP is a line segment from the centre O perpendicular to the chord AB at P.

According to the theorem which states that the perpendicular from the centre of a circle to a chord bisects the chord, the point P is the midpoint of the chord AB.

The total length of the chord AB is the sum of the lengths AP and PB:

Since $AP = PB$, we have:

Substitute the value of AP from equation (ii):

The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Final Answer:

The length of the chord is 8 cm.

Given:

A quadrilateral ABCD is drawn to circumscribe a circle with centre O.

The sides AB, BC, CD, and DA touch the circle at points P, Q, R, and S respectively.

To Prove:

$AB + CD = AD + BC$

Proof:

We know that the lengths of tangents drawn from an external point to a circle are equal.

Considering the vertices of the quadrilateral as external points, we have:

From point A, the tangents are AP and AS.

From point B, the tangents are BP and BQ.

From point C, the tangents are CR and CQ.

From point D, the tangents are DR and DS.

Add equations (1), (2), (3), and (4):

$(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)$

Rearrange the terms on the Right Hand Side (RHS):

$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$

From the figure, we can see that the sides of the quadrilateral are formed by the sum of these tangent segments:

$AB = AP + PB$ (or $AP + BP$)

$CD = CR + RD$ (or $CR + DR$)

$AD = AS + SD$ (or $AS + DS$)

$BC = BQ + QC$ (or $BQ + CQ$)

Substitute these expressions for the sides back into the equation we obtained by adding the tangent lengths:

Thus, the sum of the pairs of opposite sides of the quadrilateral circumscribing a circle are equal.

Hence, proved.

Given:

A circle with centre O.

XY and X′Y′ are two parallel tangents to the circle, with points of contact P and Q respectively.

AB is another tangent to the circle at point C, intersecting XY at A and X′Y′ at B.

To Prove:

$\angle AOB = 90^\circ$

Construction:

Join the centre O to the points of contact P, Q, and C. Also, join OA and OB.

Proof:

Since XY and X′Y′ are parallel tangents, the line segment PQ joining their points of contact must pass through the centre O and is a diameter of the circle.

Therefore, POQ is a straight line.

Consider the tangents from point A to the circle, which are AP and AC.

We know that the lengths of tangents drawn from an external point to a circle are equal.

Now, consider $\triangle OAP$ and $\triangle OCA$.

Therefore, by SSS congruence rule, $\triangle OAP$ is congruent to $\triangle OCA$.

By CPCT (Corresponding Parts of Congruent Triangles), the corresponding angles are equal:

Similarly, consider the tangents from point B to the circle, which are BQ and BC.

Now, consider $\triangle OQB$ and $\triangle OCB$.

Therefore, by SSS congruence rule, $\triangle OQB$ is congruent to $\triangle OCB$.

By CPCT, the corresponding angles are equal:

Since POQ is a straight line, the sum of the angles around O on this line is $180^\circ$.

Substitute the equal angles from equations (1) and (2) into this equation:

Factor out 2 from the left side:

Divide both sides by 2:

From the figure, the angle $\angle AOB$ is the sum of angles $\angle COA$ and $\angle COB$:

Substitute the result from equation (3) into this:

Hence, proved that $\angle AOB = 90^\circ$.

Given:

A circle with centre O.

P is an external point.

PA and PB are two tangents drawn from P to the circle, touching the circle at points A and B respectively.

To Prove:

The angle between the tangents ($\angle APB$) and the angle subtended by the line-segment joining the points of contact at the centre ($\angle AOB$) are supplementary.

That is, $\angle APB + \angle AOB = 180^\circ$.

Proof:

Let O be the centre of the circle, and P be the external point from which two tangents PA and PB are drawn to the circle, touching it at points A and B respectively.

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OA is perpendicular to the tangent PA at the point of contact A.

This implies that the angle $\angle OAP$ is a right angle.

Similarly, the radius OB is perpendicular to the tangent PB at the point of contact B.

This implies that the angle $\angle OBP$ is a right angle.

Now, consider the quadrilateral OAPB. The sum of the interior angles of a quadrilateral is $360^\circ$.

The angles of quadrilateral OAPB are $\angle OAP$, $\angle APB$, $\angle PBO$ (or $\angle OBP$), and $\angle BOA$ (or $\angle AOB$).

So, the sum of these angles is:

Substitute the known values for $\angle OAP$ and $\angle OBP$ ($90^\circ$ each) into the equation:

$90^\circ + \angle APB + 90^\circ + \angle AOB = 360^\circ$

Combine the constant terms:

$180^\circ + \angle APB + \angle AOB = 360^\circ$

Subtract $180^\circ$ from both sides of the equation:

$\angle APB + \angle AOB = 360^\circ - 180^\circ$

$\angle APB + \angle AOB = 180^\circ$

Since the sum of the two angles $\angle APB$ and $\angle AOB$ is $180^\circ$, they are supplementary angles.

Thus, the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Hence, proved.

Given:

A parallelogram ABCD is drawn to circumscribe a circle.

Let the sides AB, BC, CD, and DA touch the circle at points P, Q, R, and S respectively.

To Prove:

ABCD is a rhombus.

Proof:

Since ABCD is a parallelogram circumscribing a circle, its sides are tangent to the circle.

We know that the lengths of tangents drawn from an external point to a circle are equal.

Considering the vertices of the parallelogram as external points, we have:

From point A, the tangents are AP and AS.

From point B, the tangents are BP and BQ.

From point C, the tangents are CR and CQ.

From point D, the tangents are DR and DS.

The sides of the parallelogram can be expressed as the sum of these tangent segments:

Add the lengths of the sides AB and CD:

$AB + CD = (AP + BP) + (CR + DR)$

Add the lengths of the sides AD and BC:

$AD + BC = (AS + DS) + (BQ + CQ)$

Substitute the equal tangent lengths from equations (1) to (4) into the sum of AD and BC:

$AD + BC = (AP + DR) + (BP + CR)$

Rearrange the terms:

$AD + BC = (AP + BP) + (DR + CR)$

From equations (5) and (7), we see this is equal to $AB + CD$.

This shows that the sum of opposite sides is equal.

Since ABCD is a parallelogram, we know that opposite sides are equal:

Substitute (10) and (11) into equation (9):

$AD + AD = AB + AB$

$2 AD = 2 AB$

From (10), (11), and (12), we have:

$AB = CD$

$AD = BC$

$AB = AD$

Combining these, we get:

$AB = BC = CD = DA$

Since ABCD is a parallelogram and all its sides are equal in length, it is a rhombus.

Hence, proved.

Given:

A triangle ABC circumscribes a circle with centre O and radius $r = 4$ cm.

The point of contact on BC is D, dividing BC into segments BD and DC.

$BD = 8$ cm.

$DC = 6$ cm.

To Find:

The lengths of the sides AB and AC.

Solution:

Let the circle touch the sides AB and AC at points E and F respectively.

We know that the lengths of tangents drawn from an external point to a circle are equal.

From external point B, tangents are BD and BE.

From external point C, tangents are CD and CF.

From external point A, tangents are AE and AF.

Let $AE = AF = x$ cm.

Now, we can write the lengths of the sides of the triangle ABC:

The radius of the inscribed circle is $r = OD = OE = OF = 4$ cm.

We know that the radius is perpendicular to the tangent at the point of contact. So, $OD \perp BC$, $OE \perp AB$, and $OF \perp AC$.

The area of $\triangle ABC$ can be calculated in two ways:

Method 1: Using the sum of areas of $\triangle OBC$, $\triangle OCA$, and $\triangle OAB$.

Join the vertices A, B, C to the centre O. Also, join O to the points of contact D, E, F.

The height of each of the triangles $\triangle OBC$, $\triangle OCA$, and $\triangle OAB$ from the base (side of $\triangle ABC$) to the centre O is the radius of the inscribed circle (4 cm).

Total Area($\triangle ABC$) = Area($\triangle OBC$) + Area($\triangle OCA$) + Area($\triangle OAB$)

Method 2: Using Heron's Formula.

The sides of $\triangle ABC$ are $a = BC = 14$, $b = AC = x+6$, $c = AB = x+8$.

The semi-perimeter $s$ of $\triangle ABC$ is:

Now calculate $s-a$, $s-b$, and $s-c$:

According to Heron's formula, the Area($\triangle ABC$) is:

Equate the two expressions for the Area($\triangle ABC$) from (1) and (2):

$4x + 56 = \sqrt{48x(x+14)}$

Factor out 4 from the left side:

$4(x + 14) = \sqrt{48x(x+14)}$

Square both sides of the equation:

$(4(x + 14))^2 = (\sqrt{48x(x+14)})^2$

$16(x + 14)^2 = 48x(x+14)$

$16(x + 14)(x + 14) = 48x(x+14)$

Since x represents a length, $x \geq 0$. For a valid triangle, $x+14 > 0$. So we can divide both sides by $16(x+14)$:

$\frac{\cancel{16}\cancel{(x + 14)}(x + 14)}{\cancel{16}\cancel{(x + 14)}} = \frac{\cancel{48}^{3}x\cancel{(x+14)}}{\cancel{16}\cancel{(x + 14)}}$

$x + 14 = 3x$

Subtract $x$ from both sides:

$14 = 3x - x$

$14 = 2x$

Divide both sides by 2:

$x = \frac{14}{2}$

$x = 7$

Now we can find the lengths of sides AB and AC:

Final Answer:

The lengths of the sides AB and AC are $15$ cm and $13$ cm respectively.

Let the quadrilateral be ABCD, circumscribing a circle with centre O. Let the sides AB, BC, CD, and DA touch the circle at points P, Q, R, and S respectively.

Given:

ABCD is a quadrilateral circumscribing a circle with centre O. The sides AB, BC, CD, DA touch the circle at P, Q, R, S respectively.

To Prove:

Opposite sides subtend supplementary angles at the centre O. That is, $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.

Construction Required:

Join OA, OB, OC, OD. Also join OP, OQ, OR, OS.

Proof:

Consider the triangles $\triangle APO$ and $\triangle ASO$.

We have:

Therefore, $\triangle APO \cong \triangle ASO$ by the SSS congruence rule.

By Corresponding Parts of Congruent Triangles (CPCT), we have:

Let $\angle AOP = \angle AOS = x^\circ$.

Similarly, considering the pairs of triangles formed by the centre O and the points of contact from vertices B, C, and D, we can prove their congruence:

$\triangle BPO \cong \triangle BQO$

This implies:

Let $\angle BOP = \angle BOQ = y^\circ$.

$\triangle CQO \cong \triangle CRO$

This implies:

Let $\angle COQ = \angle COR = z^\circ$.

$\triangle DRO \cong \triangle DSO$

This implies:

Let $\angle DOR = \angle DOS = w^\circ$.

The sum of all angles around the centre O is $360^\circ$.

So, $\angle AOP + \angle AOS + \angle BOP + \angle BOQ + \angle COQ + \angle COR + \angle DOR + \angle DOS = 360^\circ$.

Substitute the angles in terms of $x, y, z, w$:

$x + x + y + y + z + z + w + w = 360^\circ$

$2x + 2y + 2z + 2w = 360^\circ$

Factor out 2:

$2(x + y + z + w) = 360^\circ$

Divide by 2:

Now, consider the angles subtended by the opposite sides AB and CD at the centre:

$\angle AOB = \angle AOP + \angle BOP = x + y$

$\angle COD = \angle COQ + \angle DOR = z + w$

The sum of these angles is:

$\angle AOB + \angle COD = (x + y) + (z + w) = x + y + z + w$

From equation (i), we know that $x + y + z + w = 180^\circ$.

Therefore,

Thus, the angles subtended by opposite sides AB and CD at the centre are supplementary.

Similarly, consider the angles subtended by the opposite sides BC and DA at the centre:

$\angle BOC = \angle BOQ + \angle COQ = y + z$

$\angle DOA = \angle DOS + \angle AOS = w + x$

The sum of these angles is:

$\angle BOC + \angle DOA = (y + z) + (w + x) = x + y + z + w$

From equation (i), we know that $x + y + z + w = 180^\circ$.

Therefore,

Thus, the angles subtended by opposite sides BC and DA at the centre are supplementary.

Hence, proved.