# Theorems and Circle Properties in Geometry

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## Theorems

**124**

#### **CHAPTER 8**

**Theorem 8.1**

**Statement:** In an obtuse-angled **triangle**, the square on the side opposite to the obtuse **angle** is equal to the sum of the squares on the sides containing the obtuse angle together with twice the rectangle contained by the sides, and the projection on it of the other.

**Given:** In ∆ABC, m∠C > 90°. Let us denote mAB = c, mAC = b and mAB = c. MAD = h is perpendicular from A to BC (produced) so that mCD = p is the projection of AC on BC.

**To Prove:**

(mAB)² = (mAC)² + (mBC)² + 2(mBC)(mCD)

Or c² = a² + b² + 2ap

**Proof:**

| Statements | Reasons |

|—|—|

| ∆ADC is a right-angled triangle | Given |

| ∴ (mAC)² = (mAD)² + (mCD)² | Pythagoras **theorem** |

| ∆ADB is a right-angled triangle | Given |

| ∴ (mAB)² = (mAD)² + (mBD)² | Pythagoras theorem |

| (mAB)² = (mAD)² + (mBC + mCD)² | mBC = mBD + mDC |

| (mAB)² = (mAD)² + (mBC)² + 2(mBC)(mCD) + (mCD)² | (x + y)² = x² + 2xy + y² |

| (mAB)² = (mAC)² + (mBC)² + 2(mBC)(mCD) | Using Equation (1) |

**Theorem 8.2:** In any triangle, the square on the side opposite to the acute angle is equal to the sum of the squares on the sides containing that acute angle diminished by twice the rectangle contained by one of those sides and the projection on it of the other.

**(Figure A: acute-angled triangle, Figure B: obtuse-angled triangle)**

**Given:** In ∆ABC, m∠C < 90°. Let us denote mAB = c, mAC = b and mAB = c. MAD = h is perpendicular from A to BC (produced) so that mCD = p is the projection of AC on BC.

**To Prove:**

(mAB)² = (mBC)² + (mCA)² – 2(mBC)(mCD)

Or c² = a² + b² – 2ap

**Proof:**

| Statements | Reasons |

|—|—|

| ∆ADC is a right-angled triangle | Given |

| ∴ (mAC)² = (mAD)² + (mCD)² | Pythagoras theorem |

| ∆ADB is a right-angled triangle | Given |

| ∴ (mAB)² = (mAD)² + (mBD)² | Pythagoras theorem |

| (mAB)² = (mAD)² + (mBC – mCD)² | mBD = mBC – mDC |

| (mAB)² = (mAD)² + (mBC)² – 2(mBC)(mCD) + (mCD)² | (x – y)² = x² – 2xy + y² |

| (mAB)² = (mAC)² + (mBC)² – 2(mBC)(mCD) | Using Equation (1) |

**Theorem 8.3**

**Statement:** In any triangle, the sum of the square of any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side.

**(Figure 1: acute-angled triangle, Figure 2: obtuse-angled triangle)**

**Given:** In ∆ABC, AE is the median drawn from the vertex A, then mBE = mEC.

**To Prove:**

(mAB)² + (mAC)² = 2(mEB)² + 2(mAE)²

**Proof:**

| Statements | Reasons |

|—|—|

| ∆ADE is a right-angled triangle | Given |

| ∴ (mAE)² = (mAD)² + (mED)² | Pythagoras theorem |

| ∆ADC is a right-angled triangle | Given |

| ∴ (mAC)² = (mAD)² + (mCD)² | Pythagoras theorem |

| ∆ADB is a right-angled triangle | Given |

| ∴ (mAB)² = (mAD)² + (mBD)² | Pythagoras theorem |

| (mAB)² + (mAC)² = (mAD)² + (mBD)² + (mAD)² + (mCD)² | Adding equations (2) and (3) |

| (mAB)² + (mAC)² = 2(mAD)² + (mBE + mED)² + (mEC – mED)² | mBD = mBE + mED, mCD = mEC – mED |

| (mAB)² + (mAC)² = 2(mAD)² + (mBE)² + 2(mBE)(mED) + (mED)² + (mEC)² – 2(mEC)(mED) + (mED)² | (x + y)² = x² + 2xy + y², (x – y)² = x² – 2xy + y² |

| (mAB)² + (mAC)² = 2(mAD)² + (mBE)² + 2(mBE)(mED) + (mED)² + (mBE)² – 2(mBE)(mED) + (mED)² | Since mBE = mEC |

| (mAB)² + (mAC)² = 2(mAD)² + 2(mBE)² + 2(mED)² | |

| (mAB)² + (mAC)² = 2((mAD)² + (mED)²) + 2(mBE)² | |

| (mAB)² + (mAC)² = 2(mAE)² + 2(mBE)² | Using Equation (1) |

**Circle:** A **circle** is a set of points in a plane which are equidistant from a fixed **point** of the plane. The fixed point is called the **center** of the **circle**.

**Radius of the Circle:** The **radius** of the circle is a straight line drawn from the center to the boundary line or circumference. The plural of the word radius is radii. (OG)

**Circumference:** The circumference of a circle is the boundary line or perimeter of the circle.

**Chord:** A line segment whose endpoints are any two points of a circle is called a **chord** of the circle. (AB)

**Diameter of the circle:** A **chord** passing through the center of the circle is called the **diameter** of the circle. (EF)

**Tangent to a circle:** A line which intersects a circle at one point only and none of its points lie in the interior of the circle is called a **tangent** to the circle. (XY)

**Secant to a circle:** If there are two distinct points common between a line and a circle, the line is called a **secant** to the circle. If there are no points common between a line and a circle, we say that the line and the circle do not intersect. (CD)

#### **CHAPTER 9**

**Theorem 9.1:** One and only one circle can pass through three non-collinear points.

**Given:** A, B, and C are three non-collinear points.

**To Prove:** One and only one circle can pass through the points A, B, and C, i.E., OA = OB = OC.

**Construction:** Join B to A and C. Draw perpendicular bisectors XD and YE of AB and BC, respectively, which intersect at O.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAD ≅ ∆OBD | |

| ∠ODA ≅ ∠ODB | XD ⊥ AB (Construction) |

| AD ≅ BD | XD is the bisector of AB |

| OD ≅ OD | Common |

| ∴ ∆OAD ≅ ∆OBD | SAS Postulate |

| ∴ OA ≅ OB | Corresponding sides of **congruent** triangles |

| Now in ∆OBE ≅ ∆OCE | |

| ∠OEB ≅ ∠OEC | YE ⊥ BC |

| BE ≅ EC | YE is the bisector of BC |

| OE ≅ OE | Common |

| ∴ ∆OBE ≅ ∆OCE | SAS Postulate |

| ∴ OB ≅ OC | Corresponding sides of congruent triangles |

| OA ≅ OB ≅ OC | From equations (1) & (2) / Transitive property |

It means that O is the only one point which is equidistant from the three points A, B, C. Therefore, one and only one circle with center O can pass through three non-collinear points, i.E., radius mOA = mOB = mOC.

**Theorem 9.2:** A straight line drawn from the center of a circle to bisect a chord (which is not a diameter) is perpendicular to the chord.

**Given:** A circle with center O, AB is a chord of the circle, N is the midpoint of AB which is joined to O.

**To Prove:** ON ⊥ AB

**Construction:** Join O to A and B.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAN ≅ ∆OBN | |

| OA ≅ OB | Radial segments |

| AN ≅ BN | N is the midpoint of AB |

| ON ≅ ON | Common |

| ∴ ∆OAN ≅ ∆OBN | SSS Postulate |

| ∴ ∠1 ≅ ∠2 | Corresponding angles of congruent triangles |

| But ∠1 + ∠2 = 180° | Supplementary angles postulate |

| ∠1 + ∠1 = 180° | ∠1 ≅ ∠2 |

| 2∠1 = 180° | |

| ∠1 = 90° | |

| ∴ ∠1 = ∠2 = 90° | |

| Hence ON ⊥ AB | Proved ∠1 = ∠2 |

**Theorem 9.3:** The perpendicular from the center of a circle on a chord bisects it.

**Given:** A circle with center O, AB is a chord and ON ⊥ AB.

**To Prove:** N is the midpoint of AB, i.E., AN ≅ BN.

**Construction:** Join O to A and B.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAN ≅ ∆OBN | |

| OA ≅ OB | Radial segments |

| ∠ONA ≅ ∠ONB | Given / Right angles |

| ON ≅ ON | Common |

| ∴ ∆OAN ≅ ∆OBN | HS Postulate |

| ∴ AN ≅ BN | Corresponding sides of congruent triangles |

**Theorem 9.4:** If two chords of a circle are congruent, then they will be equidistant from the center.

**Given:** AB and CD are two congruent chords of the circle with center O, i.E., mAB ≅ mCD.

**To Prove:** AB and CD are equidistant from the center O.

**Construction:** Join O to A and C. Also, draw perpendiculars OE and OF on the given chords AB and CD, respectively.

**Proof:**

| Statements | Reasons |

|—|—|

| mAE ≅ mEB and ∠OEA = 90° | OE is the perpendicular bisector of AB |

| mCF ≅ mFD and ∠OFC = 90° | OF is the perpendicular bisector of CD |

| mAB ≅ mCD | Given |

| mAE + mEB ≅ mCF + mFD | |

| mAE + mAE ≅ mCF + mCF | mAE ≅ mEB, mCF ≅ mFD |

| 2mAE ≅ 2mCF | |

| mAE ≅ mCF | |

| In ∆OAE ≅ ∆OCF | |

| OA ≅ OC | Radial segments |

| ∠OEA ≅ ∠OFC | Construction |

| AE ≅ CF | Proved |

| ∴ ∆OAE ≅ ∆OCF | HS Postulate |

| ∴ mOE ≅ mOF | Corresponding sides of congruent triangles |

**Theorem 9.5:** If two chords of a circle which are equidistant from the center are congruent.

**Given:** AB and CD are chords of the circle that are equidistant from center O.

**To Prove:** mAB ≅ mCD

**Construction:** Join O to A and C. Also, draw perpendicular bisectors OE and OF on the given chords AB and CD, respectively.

**Proof:**

| Statements | Reasons |

|—|—|

| mAE ≅ mEB and ∠OEA = 90° | OE is the perpendicular bisector of AB |

| mCF ≅ mFD and ∠OFC = 90° | OF is the perpendicular bisector of CD |

| In ∆OAE ≅ ∆OCF | |

| OA ≅ OC | Radial segments |

| mOE ≅ mOF | Given |

| ∠OEA ≅ ∠OFC | Construction |

| ∴ ∆OAE ≅ ∆OCF | HS Postulate |

| ∴ AE ≅ CF | Corresponding sides of congruent triangles |

| ∴ mAE ≅ mCF | |

| 2mAE ≅ 2mCF | Multiply both sides by 2 |

| mAE + mAE ≅ mCF + mCF | |

| mAE + mEB ≅ mCF + mFD | mAE ≅ mEB, mCF ≅ mFD |

| mAB ≅ mCD | |

#### **CHAPTER 10**

**Theorem 10.1:** If a line is drawn perpendicular to a radial segment of a circle at its outer endpoint, it is tangent to the circle at that point.

**Given:** OC is a radial segment of the circle with center O. The perpendicular line AB is at the outer endpoint of the circle C, i.E., AB ⊥ OC.

**To Prove:** AB is tangent to the circle.

**Construction:** Take any point D on AB other than C. Join O to D.

**Proof:**

| Statements | Reasons |

|—|—|

| OC ⊥ AB | Given |

| So the ∆OCD is a right-angled triangle | ∠OCB = 90° |

| ∴ ∠1 and ∠2 are acute angles | i.E., ∠1 + ∠2 = 90° |

| Thus ∠C > ∠2 | ∠C is right and ∠2 is acute |

| ∴ mOD > mOC | In a triangle, the greater angle has a greater opposite side |

| mOD is greater than the radial segment mOC | |

| Thus D is exterior to the circle | |

| Hence AB meets the circle at only one point, which is C. | |

**Theorem 10.2:** The tangent to a circle and the radial segment joining the point of contact and the center are perpendicular to each other.

**Given:** AB is tangent to the circle at point C, OC is a radial segment which is obtained by joining the center O and the point of contact of the tangent AB.

**To Prove:** AB ⊥ OC

**Construction:** Take any point D on AB other than C. Join O to D.

**Proof:**

| Statements | Reasons |

|—|—|

| C is the only point common to the circle and the tangent AB | AB is tangent to the circle at point C |

| ∴ D is an exterior point of the circle | Construction (Except for C, every point of AB is outside the circle) |

| It means, mOD > mOC | |

| mOC is the shortest distance between the center O and the line AB | ∠C is right (In a triangle, the greater angle has the greater side opposite to it) |

| ∴ AB ⊥ OC | By definition of shortest distance |

**Theorem 10.3:** Two tangents drawn to a circle from a point outside it are equal in length.

**Given:** A circle with center O. A is any point outside the circle. AB and AC are two tangents from point A.

**To Prove:** mAB ≅ mAC

**Construction:** Join O to A, B, and C.

**Proof:**

| Statements | Reasons |

|—|—|

| ∆AOB ≅ ∆AOC | |

| AO ≅ AO | Common |

| OB ≅ OC | Radial segments |

| ∠ABO ≅ ∠ACO | Right angles |

| ∴ ∆AOB ≅ ∆AOC | HS ≅ HS |

| AB ≅ AC | Corresponding sides of congruent triangles |

**Theorem 10.4:** If two circles touch externally, the distance between their centers is equal to the sum of their radii.

**Given:** Two circles with centers O and O’ which are touching each other externally at point A. OA and O’A are the radial segments of the circles.

**To Prove:** mOO’ = mOA + mAO’

**Construction:** Draw a common tangent BC at the point A which is the common point of the two circles.

**Proof:**

| Statements | Reasons |

|—|—|

| ∠BAO = 90° | BC is tangent to the circle at point A, i.E., OA ⊥ BC |

| ∠BAO’ = 90° | BC is tangent to the circle at point A, i.E., O’A ⊥ BC |

| ∠BAO + ∠BAO’ = 90° + 90° | |

| ∠BAO + ∠BAO’ = 180° | |

| ∴ OA and AO’ are opposite rays | ∠BAO and ∠BAO’ are supplementary angles with a common vertex A |

| Thus O, A, and O’ are three different collinear points | |

| mOO’ = mOA + mAO’ | Straight line Postulate |

**Theorem 10.5:** If two circles touch internally, the distance between their centers is the difference between their radii.

**Given:** Two circles with centers O and O’ which are touching each other internally at point A. OA and O’A are the radial segments of the circles.

**To Prove:** mOO’ = mOA – mO’A

**Construction:** Draw a common tangent BC at the point A which is the common point of the two circles.

**Proof:**

| Statements | Reasons |

|—|—|

| ∠BAO = 90° | BC is tangent to the circle at point A, i.E., OA ⊥ BC |

| ∠BAO’ = 90° | BC is tangent to the circle at point A, i.E., O’A ⊥ BC |

| ∠BAO = ∠BAO’ = 90° | |

| Thus O, A, and O’ are three different collinear points | ∠BAO and ∠BAO’ are right angles with a common vertex A |

| mOA = mOO’ + mO’A | A, O, and O’ lie on the same straight line |

| mOA – mO’A = mOO’ | From the law of equation |

| mOO’ = mOA – mO’A | |

**Arc of the Circle:** Any portion or part of the circle is called an **arc** of the circle.

**Major Arc:** An arc which is greater than a semi-circle is called a major arc.

**Central Angle:** An angle subtended by an arc at the center of a circle is called the central angle of the arc.

#### **CHAPTER 11**

**Theorem 11.1 (a):** If two arcs of a circle are congruent, then the corresponding chords are equal.

**Given:** A circle with center O. AEB and CFD are congruent arcs, i.E., AEB ≅ CFD.

**To Prove:** mAB ≅ mCD

**Construction:** Join O to A, B, C, and D, respectively, and label the central angles ∠1 and ∠2.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAB ≅ ∆OCD | |

| OA ≅ OC | In the correspondence |

| ∠1 ≅ ∠2 | Central angles of two congruent arcs |

| OB ≅ OD | Radial segments |

| ∴ ∆OAB ≅ ∆OCD | SAS Postulate |

| ∴ AB ≅ CD | Corresponding sides of congruent triangles |

**Theorem 11.1 (b):** If two arcs of congruent circles are congruent, then the corresponding chords are equal.

**Given:** Two congruent circles with centers O and O’, respectively. AEB and CFD are congruent arcs, i.E., AEB ≅ CFD.

**To Prove:** mAB ≅ mCD

**Construction:** Join O to A, B and O’ to C and D, respectively, and label the central angles ∠1 and ∠2.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAB ≅ ∆O’CD | |

| OA ≅ O’C | Radii of congruent circles |

| ∠1 ≅ ∠2 | Central angles of two congruent arcs |

| OB ≅ O’D | Radii of congruent circles |

| ∴ ∆OAB ≅ ∆O’CD | SAS Postulate |

| ∴ AB ≅ CD | Corresponding sides of congruent triangles |

**Theorem 11.2 (a):** If two chords of a circle are congruent, then the corresponding minor arcs are congruent.

**Given:** A circle with center O. AB and CD are congruent chords, i.E., mAB ≅ mCD.

**To Prove:** AEB ≅ CFD

**Construction:** Join O to A, B, C, and D, respectively, and label the central angles ∠1 and ∠2.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAB ≅ ∆OCD | |

| OA ≅ OC | In the correspondence |

| AB ≅ CD | Given |

| OB ≅ OD | Radial segments |

| ∴ ∆OAB ≅ ∆OCD | SSS Postulate |

| ∴ ∠1 ≅ ∠2 | Corresponding angles of congruent triangles |

| ∴ AEB ≅ CFD | Congruent arcs have congruent central angles |

**Theorem 11.2 (b):** If two chords of a circle are congruent, then the corresponding major arcs are congruent.

**Given:** A circle with center O. AB and CD are congruent chords, i.E., mAB ≅ mCD.

**To Prove:** AFB ≅ CED

**Construction:** Join O to A, B, C, and D, respectively, and label the central angles ∠1 and ∠2.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAB ≅ ∆OCD | |

| OA ≅ OC | In the correspondence |

| AB ≅ CD | Given |

| OB ≅ OD | Radial segments |

| ∴ ∆OAB ≅ ∆OCD | SSS Postulate |

| ∴ ∠1 ≅ ∠2 | Corresponding angles of congruent triangles |

| ∴ AEB ≅ CFD | Congruent arcs have congruent central angles |

| mAFB ≅ 360° – mAED | Total angle at a point = 360°, i.E., mAED + mAFB = 360° |

| mCED ≅ 360° – mCFD | Total angle at a point = 360°, i.E., mCFD + mCED = 360° |

| ∴ mAFB ≅ mCED | Transitive property |

**Theorem 11.2 (c):** If two chords of congruent circles are congruent, then the corresponding minor arcs are congruent.

**Given:** Two congruent circles with centers O and O’, respectively. AB and CD are congruent chords, i.E., mAB ≅ mCD.

**To Prove:** AEB ≅ CFD

**Construction:** Join O to A, B and O’ to C and D, respectively, and label the central angles ∠1 and ∠2.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAB ≅ ∆O’CD | |

| OA ≅ O’C | Radii of congruent circles |

| AB ≅ CD | Given |

| OB ≅ O’D | Radii of congruent circles |

| ∴ ∆OAB ≅ ∆O’CD | SSS Postulate |

| ∴ ∠1 ≅ ∠2 | Corresponding angles of congruent triangles |

| ∴ mAED ≅ mCFD | Congruent arcs have congruent central angles |

**Theorem 11.2 (d):** If two chords of congruent circles are congruent, then the corresponding major arcs are congruent.

**Given:** Two congruent circles with centers O and O’, respectively. AB and CD are congruent chords, i.E., mAB ≅ mCD.

**To Prove:** AGB ≅ CHD

**Construction:** Join O to A, B and O’ to C and D, respectively, and label the central angles ∠1 and ∠2.

**Proof:**

| Statements | Reasons |

|—|—|

| In ∆OAB ≅ ∆O’CD | |

| OA ≅ O’C | Radii of congruent circles |

| AB ≅ CD | Given |

| OB ≅ O’D | Radii of congruent circles |

| ∴ ∆OAB ≅ ∆O’CD | SSS Postulate |

| ∴ ∠1 ≅ ∠2 | Corresponding angles of congruent triangles |

| ∴ mAED ≅ mCFD | Congruent arcs have congruent central angles |

| mAGB ≅ 360° – mAED | Total angle at a point = 360°, i.E., mAED + mAGB = 360° |

| mCHD ≅ 360° – mCFD | Total angle at a point = 360°, i.E., mCFD + mCHD = 360° |

| ∴ mAGB ≅ mCHD | Transitive property |

**Theorem 11.2 (e):** If two chords (diameters) of a circle are congruent, then the corresponding minor/major arcs are congruent.

**Given:** A circle with center O. AB and CD are diameters, i.E., mAB ≅ mCD.

**To Prove:** AFB ≅ AEB ≅ CED ≅ CFD

**Proof:**

| Statements | Reasons |

|—|—|

| mAB ≅ mCD | Diameters of the same circle |

| Then the length of the minor arc = length of the major arc | |

| i.E., mAED ≅ mAFB ≅ 180° | Congruent arcs have congruent central angles |

| Similarly, mCFD ≅ mCED ≅ 180° | Diameter is a straight line which has a supplementary angle |

| ∴ mAFB ≅ mAED ≅ mCED ≅ mCFD | Transitive property |

**Theorem 11.2 (f):** If two chords (diameters) of two congruent circles are congruent, then the corresponding minor/major arcs are congruent.

**Given:** Two circles with centers O and O’. AB and CD are diameters, i