Topics
Similarity
- Similarity of Triangles
- Properties of Ratios of Areas of Two Triangles
- Basic Proportionality Theorem (Thales Theorem)
- Converse of Basic Proportionality Theorem
- Property of an Angle Bisector of a Triangle
- Property of Three Parallel Lines and Their Transversals
- Similar Triangles
- Criteria for Similarity of Triangles
- Areas of Similar Triangles
Pythagoras Theorem
- Pythagoras Theorem
- Pythagorean Triplet
- Property of 30°- 60°- 90° Triangle Theorem
- Property of 45°- 45°- 90° Triangle Theorem
- Similarity in Right Angled Triangles
- Theorem of Geometric Mean
- Right-angled Triangles and Pythagoras Property
- Converse of Pythagoras Theorem
- Application of Pythagoras Theorem in Acute Angle and Obtuse Angle
- Apollonius Theorem
Circle
- Concept of Circle
- Circles Passing Through One, Two, Three Points
- Secant and Tangent
- Tangent to a Circle
- Converse of Tangent Theorem
- Tangent Segment Theorem
- Touching Circles
- Theorem of Touching Circles
- Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers
- Introduction to an Arc
- Congruence of Arcs
- Property of Sum of Measures of Arcs
- Inscribed Angle
- Intercepted Arc
- Inscribed Angle Theorem
- Corollaries of Inscribed Angle Theorem
- Cyclic Quadrilateral
- Theorem: Opposite angles of a cyclic quadrilateral are supplementary.
- Corollary of Cyclic Quadrilateral Theorem
- Converse: If a Pair of Opposite Angles of a Quadrilateral is Supplementary, Then the Quadrilateral is Cyclic.
- Converse of Cyclic Quadrilateral Theorem
- Theorem of Angle Between Tangent and Secant
- Converse of Theorem of the Angle Between Tangent and Secant
- Theorem of Internal Division of Chords
- Theorem of External Division of Chords
- Tangent Secant Segments Theorem
- Tangent - Secant Theorem
- Angle Subtended by the Arc to the Point on the Circle
- Angle Subtended by the Arc to the Centre
- Number of Tangents from a Point on a Circle
Geometric Constructions
- Basic Geometric Constructions
- Division of a Line Segment
- Construction of Similar Triangle
- Construction of a Tangent to the Circle at a Point on the Circle
- To Construct Tangents to a Circle from a Point Outside the Circle.
Co-ordinate Geometry
- Coordinate Geometry
- Distance Formula
- Intercepts Made by a Line
- Division of a Line Segment
- Section Formula
- The Mid-point of a Line Segment (Mid-point Formula)
- Centroid Formula
- Slope of a Line
- General Equation of a Line
- Standard Forms of Equation of a Line
Trigonometry
Mensuration
- Conversion of Solid from One Shape to Another
- Euler's Formula
- Concept of Surface Area, Volume, and Capacity
- Surface Area and Volume of Three Dimensional Figures
- Surface Area and Volume of Different Combination of Solid Figures
- Frustum of a Cone
- Sector of a Circle
- Area of a Sector of a Circle
- Length of an Arc
- Segment of a Circle
- Area of a Segment
- Circumference of a Circle
- Areas of Sector and Segment of a Circle
Theorem of Angle Between Tangent and Secant
If an angle has its vertex on the circle, its one side touches the circle and the other intersects the circle in one more point, then the measure of the angle is half the measure of its intercepted arc.
Given: Let ∠ ABC be an angle, where vertex B lies on a circle with centre M.
Its side BC touches the circle at B and side BA intersects the circle at A. Arc ADB is intercepted by ∠ABC.
To prove: ∠ ABC = 1/2 m(arc ADB)
Proof: Consider three cases.
(1) In the above figure (i), the centre M lies on the arm BA of ∠ABC,
∠ABC = ∠ MBC = 90° ..... tangent theorem (I)
arc ADB is a semicircle.
∴ m(arc ADB) = 180° ..... definition of measure of arc (II)
From (I) and (II)
∠ABC = 1/2 m(arc ADB)
(2) In the above figure (ii) centre M lies in the exterior of ∠ABC,
Draw radii MA and MB.
Now, ∠MBA = ∠MAB ..... isosceles triangle theorem
∠ MBC = 90° ..... tangent theorem..... (I)
let ∠ MBA = ∠ MAB = x and ∠ ABC = y.
∠ AMB = 180 - (x + x) = 180 - 2x
∠ MBC = ∠ MBA + ∠ABC = x + y
∴ x + y = 90°
∴ 2x + 2y = 180°
In Δ AMB, 2x + ∠AMB = 180°
∴ 2x + 2y = 2x + ∠ AMB
∴ 2y = ∠ AMB
∴ y = ∠ ABC = 1
2 ∠ AMB = 1
2 m(arc ADB)
