Topics
Number Systems
Number Systems
Polynomials
Algebra
Coordinate Geometry
Linear Equations in Two Variables
Geometry
Coordinate Geometry
Introduction to Euclid’S Geometry
Mensuration
Statistics and Probability
Lines and Angles
- Introduction to Lines and Angles
- Basic Terms and Definitions
- Intersecting Lines and Non-intersecting Lines
- Introduction to Parallel Lines
- Pairs of Angles
- Parallel Lines and a Transversal
- Angle Sum Property of a Triangle
Triangles
- Concept of Triangles
- Congruence of Triangles
- Criteria for Congruence of Triangles
- Properties of a Triangle
- Some More Criteria for Congruence of Triangles
- Inequalities in a Triangle
Quadrilaterals
- Concept of Quadrilaterals
- Properties of a Quadrilateral
- Types of Quadrilaterals
- Another Condition for a Quadrilateral to Be a Parallelogram
- Theorem of Midpoints of Two Sides of a Triangle
- Property: The Opposite Sides of a Parallelogram Are of Equal Length.
- Theorem: A Diagonal of a Parallelogram Divides It into Two Congruent Triangles.
- Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram.
- Property: The Opposite Angles of a Parallelogram Are of Equal Measure.
- Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram.
- Property: The diagonals of a parallelogram bisect each other. (at the point of their intersection)
- Theorem : If the Diagonals of a Quadrilateral Bisect Each Other, Then It is a Parallelogram
Circles
Areas - Heron’S Formula
Surface Areas and Volumes
Statistics
Algebraic Expressions
Algebraic Identities
Area
Constructions
- Introduction of Constructions
- Basic Constructions
- Some Constructions of Triangles
Probability
Theorem
Theorem: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Proof: In quadrilateral ABCD
∠A = ∠D
`=>` ∠A +∠B +∠C + ∠D
Now, ∠A +∠B +∠C + ∠D = 360° (angle sum property of quadrilateral)
`=>` 2(∠A +∠B ) = 360°
`=> ` ∠A +∠B = 180°
`therefore` ∠A +∠B = ∠C + ∠D = 180°
Line AB intersects AD and BC at A and B respectively.
Such that ∠A +∠B = 180°
`therefore` AD || BC (Sum of consecutive interior angle is 180° ) ...(1)
∠A +∠B = 180°
∠A +∠D = 180° (∠B= ∠D)
`therefore` AB || CD ..(2)
From (1) and (2), we get
AB || CD and AD || BC
`therefore` ABCD is a parallelogram.
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