Given: A quadrilateral $ABCD$ circumscribing a circle with center $O$. The sides $AB, BC, CD,$ and $DA$ touch the circle at points $P, Q, R,$ and $S$ respectively.

To Prove: The opposite sides subtend supplementary angles at the center $O$. That is, $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.

Step 1: Joining the points of contact to the center.
Join the center $O$ to the points of contact $P, Q, R,$ and $S$. Also, join $OA, OB, OC,$ and $OD$.

Step 2: Identifying congruent triangles.
Consider $\triangle OAP$ and $\triangle OAS$:

• $AP = AS$ [Tangents drawn from an external point to a circle are equal in length]
• $OA = OA$ [Common side]
• $OP = OS$ [Radii of the same circle]

Step 3: Applying the same logic to other pairs.
Similarly, we can prove:

• $\triangle OBP \cong \triangle OBQ \implies \angle 3 = \angle 4$
• $\triangle OCQ \cong \triangle OCR \implies \angle 5 = \angle 6$
• $\triangle ODR \cong \triangle ODS \implies \angle 7 = \angle 8$

Step 4: Summing the angles around the center.
The sum of all angles around the center $O$ is $360^\circ$: $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$

Step 5: Substituting the equal angles.
Since $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, and $\angle 7 = \angle 8$, we substitute these into the equation: $2\angle 2 + 2\angle 3 + 2\angle 6 + 2\angle 7 = 360^\circ$ Dividing by 2: $\angle 2 + \angle 3 + \angle 6 + \angle 7 = 180^\circ$

Step 6: Grouping the angles.
Rearranging the terms: $(\angle 2 + \angle 3) + (\angle 6 + \angle 7) = 180^\circ$ From the figure, $\angle 2 + \angle 3 = \angle AOB$ and $\angle 6 + \angle 7 = \angle COD$. Thus, $\angle AOB + \angle COD = 180^\circ$.

Similarly, it can be shown that $\angle BOC + \angle DOA = 180^\circ$.

Final Answer: The opposite sides of the quadrilateral subtend supplementary angles at the center of the circle, as $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.

Given: A quadrilateral $ABCD$ circumscribing a circle with center $O$. The circle touches the sides $AB$, $BC$, $CD$, and $DA$ at points $P$, $Q$, $R$, and $S$ respectively.

To Prove: $AB + CD = AD + BC$

Step 1: Applying the Tangent Theorem

We use the theorem: The lengths of tangents drawn from an external point to a circle are equal.

Considering the external points $A, B, C,$ and $D$:

1. From point $A$: $AP = AS$ --- (Equation 1)

2. From point $B$: $BP = BQ$ --- (Equation 2)

3. From point $C$: $CQ = CR$ --- (Equation 3)

4. From point $D$: $DR = DS$ --- (Equation 4)

Step 2: Summing the Equations

Adding Equations 1, 2, 3, and 4 together:

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

Step 3: Rearranging the Terms

Group the segments that form the sides of the quadrilateral:

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

Step 4: Substituting Side Lengths

Based on the geometry of the quadrilateral:

$AP + BP = AB$

$CQ + DR = CD$

$AS + DS = AD$

$BQ + CR = BC$

Substituting these into the equation from Step 3:

$AB + CD = AD + BC$

Conclusion:

Since the sum of the lengths of opposite sides is equal, the identity is proven.

Final Answer: Hence, it is proved that $AB + CD = AD + BC$.

Given:

A circle with center $O$. A point $Q$ lies outside the circle such that the distance from the center $O$ to point $Q$ is $OQ = 25$ cm. A tangent is drawn from $Q$ to the circle, touching the circle at point $P$, such that the length of the tangent $QP = 24$ cm.

To Find:

The radius of the circle, denoted as $OP$.

Step 1: Applying the Tangent-Radius Theorem

According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."

Therefore, $OP \perp QP$. This implies that $\angle OPQ = 90^\circ$.

Step 2: Identifying the Triangle

Since $\angle OPQ = 90^\circ$, the triangle $\triangle OPQ$ is a right-angled triangle, where $OQ$ is the hypotenuse, and $OP$ and $QP$ are the legs of the triangle.

Step 3: Applying 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. [Pythagorean Theorem: $a^2 + b^2 = c^2$]

Substituting the known values into the equation:

$OP^2 + QP^2 = OQ^2$

$OP^2 + (24)^2 = (25)^2$

Step 4: Algebraic Calculation

Calculate the squares of the given lengths:

$OP^2 + 576 = 625$

Isolate $OP^2$ by subtracting 576 from both sides:

$OP^2 = 625 - 576$

$OP^2 = 49$

Take the square root of both sides:

$OP = \sqrt{49}$

$OP = 7$ cm

Conclusion:

The radius of the circle $OP$ is $7$ cm.

Final Answer: 7 cm

Given:

A circle with centre $O$. $TP$ and $TQ$ are two tangents drawn from an external point $T$ to the circle. The angle between the radii at the points of contact is $\angle POQ = 110^{\circ}$.

To Find:

The measure of $\angle PTQ$.

Visual Representation:

Step 1: Identifying Geometric Properties

According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."

Therefore, $OP \perp TP$ and $OQ \perp TQ$.

This implies that $\angle OPT = 90^{\circ}$ and $\angle OQT = 90^{\circ}$.

Step 2: Analyzing the Quadrilateral

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

The sum of the angles is given by:

$\angle POQ + \angle OPT + \angle PTQ + \angle OQT = 360^{\circ}$

Step 3: Substituting Known Values

Substitute the known values into the equation:

$110^{\circ} + 90^{\circ} + \angle PTQ + 90^{\circ} = 360^{\circ}$

Step 4: Solving for $\angle PTQ$

Combine the constant terms:

$110^{\circ} + 180^{\circ} + \angle PTQ = 360^{\circ}$

$290^{\circ} + \angle PTQ = 360^{\circ}$

Subtract $290^{\circ}$ from both sides:

$\angle PTQ = 360^{\circ} - 290^{\circ}$

$\angle PTQ = 70^{\circ}$

Final Answer: The measure of $\angle PTQ$ is $70^{\circ}$.

Given:

1. A circle with center $O$.

2. Two tangents $PA$ and $PB$ drawn from an external point $P$ to the circle.

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

To Find:

The measure of $\angle POA$.

Step 1: Analyzing the Geometry of the Quadrilateral

Consider the quadrilateral $OAPB$. In this figure:

1. $OA \perp PA$ [Since the tangent at any point of a circle is perpendicular to the radius through the point of contact]. Thus, $\angle OAP = 90^{\circ}$.

2. $OB \perp PB$ [Since the tangent at any point of a circle is perpendicular to the radius through the point of contact]. Thus, $\angle OBP = 90^{\circ}$.

Step 2: Calculating the sum of angles in the quadrilateral

The sum of the interior angles of a quadrilateral is $360^{\circ}$.

$\angle OAP + \angle OBP + \angle APB + \angle AOB = 360^{\circ}$

Substituting the known values:

$90^{\circ} + 90^{\circ} + 80^{\circ} + \angle AOB = 360^{\circ}$

$260^{\circ} + \angle AOB = 360^{\circ}$

$\angle AOB = 360^{\circ} - 260^{\circ} = 100^{\circ}$

Step 3: Using Congruency to find $\angle POA$

Consider $\triangle OAP$ and $\triangle OBP$:

1. $OA = OB$ (Radii of the same circle)

2. $OP = OP$ (Common side)

3. $PA = PB$ (Tangents drawn from an external point to a circle are equal in length)

By SSS congruency criterion, $\triangle OAP \cong \triangle OBP$.

By CPCT (Corresponding Parts of Congruent Triangles), $\angle POA = \angle POB$.

Since $\angle AOB = \angle POA + \angle POB$, we have:

$\angle AOB = 2 \times \angle POA$

$100^{\circ} = 2 \times \angle POA$

$\angle POA = \frac{100^{\circ}}{2} = 50^{\circ}$

Final Answer: The value of $\angle POA$ is $50^{\circ}$.

Given: A parallelogram $ABCD$ circumscribing a circle with center $O$. The sides $AB$, $BC$, $CD$, and $DA$ touch the circle at points $P$, $Q$, $R$, and $S$ respectively.

To Prove: Parallelogram $ABCD$ is a rhombus (i.e., $AB = BC = CD = DA$).

Step 1: Applying the Tangent Theorem
We know that the lengths of tangents drawn from an external point to a circle are equal. [Theorem: Tangents from an external point to a circle are equal in length.]

From point $A$: $AP = AS$ --- (i)

From point $B$: $BP = BQ$ --- (ii)

From point $C$: $CQ = CR$ --- (iii)

From point $D$: $DR = DS$ --- (iv)

Step 2: Summing the Equations
Adding equations (i), (ii), (iii), and (iv):
$(AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CQ)$
Rearranging the terms based on the sides of the parallelogram:
$(AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CQ)$
$AB + CD = AD + BC$ --- (v)

Step 3: Utilizing Parallelogram Properties
Since $ABCD$ is a parallelogram, we know that opposite sides are equal:
$AB = CD$ and $BC = AD$ [Property of a parallelogram]

Step 4: Substitution and Simplification
Substitute $CD = AB$ and $AD = BC$ into equation (v):
$AB + AB = BC + BC$
$2AB = 2BC$
$AB = BC$

Step 5: Conclusion
Since $AB = BC$ and we already know $AB = CD$ and $BC = AD$ (opposite sides of a parallelogram), it follows that:
$AB = BC = CD = DA$.

A parallelogram with all sides equal is defined as a rhombus. Therefore, $ABCD$ is a rhombus.

Final Answer: Since all sides of the parallelogram $ABCD$ are equal ($AB = BC = CD = DA$), the parallelogram is proven to be a rhombus.

Given: A circle with center $O$. Let $XY$ be a tangent to the circle at the point of contact $P$.

To Prove: The perpendicular to the tangent $XY$ at the point of contact $P$ passes through the center $O$.

Step 1: Assumption for Proof by Contradiction
Let us assume that the perpendicular to the tangent $XY$ at point $P$ does not pass through the center $O$. Instead, let it pass through another point $O'$ such that $O'P \perp XY$.

Step 2: Utilizing the Tangent-Radius Theorem
We know from the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."
Therefore, the radius $OP$ must be perpendicular to the tangent $XY$.
Mathematically, $OP \perp XY$.
This implies that $\angle OPX = 90^\circ$. [Since $OP$ is the radius at point of contact $P$]

Step 3: Analyzing the Assumption
By our initial assumption in Step 1, we stated that $O'P \perp XY$.
This implies that $\angle O'PX = 90^\circ$.

Step 4: Logical Deduction
Comparing the results from Step 2 and Step 3:
$\angle OPX = 90^\circ$
$\angle O'PX = 90^\circ$
Therefore, $\angle OPX = \angle O'PX$.

Step 5: Conclusion
The equality $\angle OPX = \angle O'PX$ is only possible if the line $O'P$ coincides with the line $OP$.
This means that the point $O'$ must be the same as point $O$.
Thus, our assumption that the perpendicular does not pass through the center is false.

Final Answer: Hence, it is proved that the perpendicular at the point of contact to the tangent to a circle must pass through the center.

Given:

1. A circle with center $O$.

2. A point $A$ outside the circle such that the distance from the center $O$ to point $A$ is $OA = 5$ cm.

3. A tangent $AB$ from point $A$ to the circle at point $B$, where the length of the tangent $AB = 4$ cm.

To find:

The radius of the circle, denoted as $OB$.

Step 1: Applying the Tangent-Radius Theorem

According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."

Therefore, $OB \perp AB$. This implies that $\angle OBA = 90^\circ$.

Step 2: Identifying the Geometric Figure

Since $\angle OBA = 90^\circ$, the triangle $\triangle OBA$ is a right-angled triangle, where $OA$ is the hypotenuse, $OB$ is the radius (base/height), and $AB$ is the tangent.

Step 3: Applying 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. [Using Pythagoras Theorem: $Hypotenuse^2 = Base^2 + Height^2$]

$OA^2 = OB^2 + AB^2$

Step 4: Substituting the Given Values

Given $OA = 5$ cm and $AB = 4$ cm.

$(5)^2 = OB^2 + (4)^2$

Step 5: Solving for the Radius ($OB$)

$25 = OB^2 + 16$

Subtract 16 from both sides:

$OB^2 = 25 - 16$

$OB^2 = 9$

Taking the square root on both sides:

$OB = \sqrt{9}$

$OB = 3$ cm

[Since length cannot be negative, we take the positive root.]

Final Answer: The radius of the circle is 3 cm.

Given: A circle with center $O$. An external point $P$ from which two tangents $PA$ and $PB$ are drawn to the circle, touching the circle at points $A$ and $B$ respectively. The line segment $AB$ joins the points of contact.

To Prove: $\angle APB + \angle AOB = 180^\circ$ (i.e., they are supplementary).

Step 1: Identify the properties of tangents to a circle.

According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."

Therefore, $OA \perp PA$ and $OB \perp PB$.

This implies that $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.

Step 2: Consider the quadrilateral $OAPB$.

In the quadrilateral $OAPB$, the sum of the interior angles is given by the angle sum property of a quadrilateral.

$\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^\circ$ [Sum of interior angles of a quadrilateral is $360^\circ$].

Step 3: Substitute the known values into the equation.

Substitute $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ into the equation:

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

Step 4: Simplify the algebraic expression.

Combine the constant terms:

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

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

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

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

Conclusion:

Since the sum of $\angle APB$ and $\angle AOB$ is $180^\circ$, the angle between the two tangents and the angle subtended by the line segment joining the points of contact at the center are supplementary.

Final Answer: Hence, it is proved that $\angle APB + \angle AOB = 180^\circ$.

Given:

Two concentric circles with a common center $O$.

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

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

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

To Find:

The length of the chord $AB$.

Step 1: Establishing Geometric Relationships

Let $O$ be the center of the concentric circles. Let $AB$ be the chord of the larger circle that is tangent to the smaller circle at point $P$.

By the property of tangents: A tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, $OP \perp AB$. [Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.]

Step 2: Applying the Perpendicular Bisector Theorem

In the larger circle, $OP$ is a line segment from the center $O$ perpendicular to the chord $AB$.

According to the theorem: A perpendicular drawn from the center of a circle to a chord bisects the chord. [Theorem: The perpendicular from the center of a circle to a chord bisects the chord.]

Therefore, $AP = PB$.

Step 3: Calculating the Length of $AP$ using the Pythagorean Theorem

Consider the right-angled triangle $\triangle OPA$, where $\angle OPA = 90^\circ$.

Using the Pythagorean Theorem: $OA^2 = OP^2 + AP^2$.

Given $OA = R = 5$ cm and $OP = r = 3$ cm.

$5^2 = 3^2 + AP^2$

$25 = 9 + AP^2$

$AP^2 = 25 - 9$

$AP^2 = 16$

$AP = \sqrt{16} = 4$ cm.

Step 4: Determining the Total Length of the Chord $AB$

Since $AP = PB$ and $AP = 4$ cm, then $PB = 4$ cm.

The total length of the chord $AB = AP + PB$.

$AB = 4 \text{ cm} + 4 \text{ cm} = 8 \text{ cm}$.

Final Answer: The length of the chord of the larger circle is 8 cm.

Given:

A circle with center $O$ and radius $r = 4$ cm is inscribed in $\triangle ABC$. The circle touches the sides $BC$, $AC$, and $AB$ at points $D$, $E$, and $F$ respectively. The segments $BD = 8$ cm and $DC = 6$ cm.

To Find:

The lengths of sides $AB$ and $AC$.

Step 1: Applying the Tangent Properties

According to the theorem: "The lengths of tangents drawn from an external point to a circle are equal."

Let $AF = AE = x$ cm.

Since $BD = 8$ cm, then $BF = BD = 8$ cm.

Since $DC = 6$ cm, then $CE = DC = 6$ cm.

Therefore, the sides of the triangle are:

$BC = BD + DC = 8 + 6 = 14$ cm

$AB = AF + FB = x + 8$ cm

$AC = AE + EC = x + 6$ cm

Step 2: Calculating the Area of $\triangle ABC$ using Heron's Formula

The semi-perimeter $s$ is given by:

$s = \frac{AB + BC + AC}{2} = \frac{(x + 8) + 14 + (x + 6)}{2} = \frac{2x + 28}{2} = x + 14$

Area of $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$

Area $= \sqrt{(x+14)(x+14 - 14)(x+14 - (x+6))(x+14 - (x+8))}$

Area $= \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)}$

Step 3: Calculating the Area using the Incenter Method

The area of $\triangle ABC$ is the sum of the areas of $\triangle OBC$, $\triangle OCA$, and $\triangle OAB$.

Area $= \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB)$

Area $= \frac{1}{2} \times BC \times r + \frac{1}{2} \times AC \times r + \frac{1}{2} \times AB \times r$

Area $= \frac{1}{2} \times 4 \times (14 + x + 6 + x + 8) = 2 \times (2x + 28) = 4(x + 14)$

Step 4: Equating the Areas and Solving for $x$

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

Squaring both sides:

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

Divide both sides by $16(x+14)$ (since $x+14 \neq 0$):

$3x = x + 14$

$2x = 14 \implies x = 7$

Step 5: Determining Final Side Lengths

$AB = x + 8 = 7 + 8 = 15$ cm

$AC = x + 6 = 7 + 6 = 13$ cm

Final Answer: The sides are $AB = 15$ cm and $AC = 13$ cm.

Given:

1. A circle with center $O$.

2. Two parallel tangents $XY$ and $X'Y'$ touching the circle at points $P$ and $Q$ respectively.

3. A third tangent $AB$ touching the circle at point $C$, intersecting $XY$ at $A$ and $X'Y'$ at $B$.

To Prove:

$\angle AOB = 90^{\circ}$

Step 1: Construction and Identification of Congruent Triangles

Join $OC$. Consider $\triangle OPA$ and $\triangle OCA$.

In $\triangle OPA$ and $\triangle OCA$:

$OP = OC$ [Radii of the same circle]

$OA = OA$ [Common side]

$AP = AC$ [Tangents drawn from an external point $A$ to the circle are equal in length]

Therefore, $\triangle OPA \cong \triangle OCA$ [By SSS Congruence Criterion].

Consequently, $\angle POA = \angle COA$ (Equation 1) [By CPCT - Corresponding Parts of Congruent Triangles].

Step 2: Applying Congruence to the Second Set of Triangles

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

In $\triangle OQB$ and $\triangle OCB$:

$OQ = OC$ [Radii of the same circle]

$OB = OB$ [Common side]

$BQ = BC$ [Tangents drawn from an external point $B$ to the circle are equal in length]

Therefore, $\triangle OQB \cong \triangle OCB$ [By SSS Congruence Criterion].

Consequently, $\angle QOB = \angle COB$ (Equation 2) [By CPCT].

Step 3: Summing the Angles on the Straight Line

$PQ$ is a diameter of the circle because $XY \parallel X'Y'$ and the tangents are perpendicular to the diameter at the points of contact. Thus, $POQ$ is a straight line.

The sum of all angles formed at the center $O$ on the straight line $PQ$ is $180^{\circ}$:

$\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}$

Step 4: Substitution and Final Calculation

Using Equation 1 ($\angle POA = \angle COA$) and Equation 2 ($\angle QOB = \angle COB$):

$\angle COA + \angle COA + \angle COB + \angle COB = 180^{\circ}$

$2\angle COA + 2\angle COB = 180^{\circ}$

$2(\angle COA + \angle COB) = 180^{\circ}$

$\angle COA + \angle COB = \frac{180^{\circ}}{2}$

$\angle AOB = 90^{\circ}$ [Since $\angle COA + \angle COB = \angle AOB$]

Final Answer: $\angle AOB = 90^{\circ}$

Given: A circle with center $O$. Let $AB$ be a diameter of the circle. Let $PQ$ be a tangent at point $A$ and $RS$ be a tangent at point $B$.

To Prove: $PQ \parallel RS$.

Step 1: Applying the Tangent-Radius Theorem
According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."
Since $PQ$ is a tangent at $A$ and $OA$ is the radius, we have:
$OA \perp PQ \implies \angle OAP = 90^\circ$ and $\angle OAQ = 90^\circ$.
Similarly, since $RS$ is a tangent at $B$ and $OB$ is the radius, we have:
$OB \perp RS \implies \angle OBR = 90^\circ$ and $\angle OBS = 90^\circ$.

Step 2: Analyzing the Angles
Consider the lines $PQ$ and $RS$ intersected by the transversal $AB$.
From Step 1, we have:
$\angle OAP = 90^\circ$
$\angle OBS = 90^\circ$
Therefore, $\angle OAP = \angle OBS = 90^\circ$.

Step 3: Establishing Parallelism
Observe that $\angle OAP$ and $\angle OBS$ are alternate interior angles with respect to lines $PQ$ and $RS$ and transversal $AB$.
[Theorem: If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel.]
Since $\angle OAP = \angle OBS = 90^\circ$, the alternate interior angles are equal.
Thus, $PQ \parallel RS$.

Final Answer: Since the alternate interior angles formed by the transversal $AB$ with lines $PQ$ and $RS$ are equal ($90^\circ$), the tangents $PQ$ and $RS$ are parallel to each other.