#### Topics

##### Similarity

- Property of three parallel lines and their transversals
- Property of an Angle Bisector of a Triangle
- Basic Proportionality Theorem Or Thales Theorem
- Converse of Basic Proportionality Theorem
- Appolonius Theorem
- Application of Pythagoras Theorem in Acute Angle and Obtuse Angle
- Similarity
- Properties of Ratios of Areas of Two Triangles
- Similarity of Triangles
- Similar Triangles
- Similarity Triangle Theorem
- Areas of Two Similar Triangles
- Areas of Similar Triangles

##### Pythagoras Theorem

##### Circle

- Theorem of External Division of Chords
- Theorem of Internal Division of Chords
- Converse of Theorem of the Angle Between Tangent and Secant
- Theorem of Angle Between Tangent and Secant
- Converse: If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.
- Corollary of Cyclic Quadrilateral Theorem
- Theorem: Opposite angles of a cyclic quadrilateral are supplementary.
- Corollaries of Inscribed Angle Theorem
- Inscribed Angle Theorem
- Intercepted Arc
- Inscribed Angle
- Property of Sum of Measures of Arcs
- Tangent Segment Theorem
- Converse of Tangent Theorem
- Circles Passing Through One, Two, Three Points
- Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers
- Cyclic Properties
- Tangent - Secant Theorem
- Cyclic Quadrilateral
- Angle Subtended by the Arc to the Point on the Circle
- Angle Subtended by the Arc to the Centre
- Introduction to an Arc
- Touching Circles
- Number of Tangents from a Point on a Circle
- Tangent to a Circle
- Tangents and Its Properties
- Theorem - Converse of Tangent at Any Point to the Circle is Perpendicular to the Radius
- Number of Tangents from a Point on a Circle
- Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles

##### Co-ordinate Geometry

##### Geometric Constructions

- To Construct Tangents to a Circle from a Point Outside the Circle.
- Construction of Triangle If the Base, Angle Opposite to It and Either Median Altitude is Given
- Construction of Tangent Without Using Centre
- Construction of Tangents to a Circle
- Construction of Tangent to the Circle from the Point on the Circle
- Basic Geometric Constructions
- Division of a Line Segment

##### Trigonometry

##### Mensuration

#### notes

We know that through one point say 'P' we can draw many circles.

Through two point say 'P' and 'Q' again we can draw many circles.

Now take three points 'A' , 'B' and 'C'.

After drawing perpendicular bisector AB on 'PQ' and RS on 'BC', we find that OA = OB = OC.

Where O is the point where both the perpendicular bisector meet.

Now a circle can be drawn by taking O as a centre and OA as a radius. This demonstration proves the theorem that There is one and only one circle passing through three given non-collinear points.

This circle is called the circumcircle of the ∆ ABC. Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.

#### theorem

**Theorem :** There is one and only one circle passing through three given non-collinear points.

**Proof : **

**Given :** We have three non -collinear points A , B and C (As shown e fig.)**To prove :** We need to prove that there is ONE and ONLY one circle that passes through all the three points A , B and C.

We join AB and BC . Then we draw a perpendicular bisector of AB say line QP . Also draw a perpendicular bisector of BC say line RS .

Since AB and BC are not parallel , their perpendicular bisector i.e. line QP and line RS will not be parallel. Say line QP and line RS intersect at point O.

Now we know that if we take any point on the perpendicular bisector of AB , that point will be equidistant from both A and B . So, O is on line QP.

Therefore , OA = OB .....(1)

Similarly , if we take any point on the perpendicular bisector of BC , that point will be equidistance from both B and C. So, o is on line RS.

Therefore, OB = OC ....(2)

From (1) and (2) , we get

OA =OB =OC

We see that O is such a point that is equidistant from all three points A,B and C. So, we can draw a circle with centre O and radius OA that will passed through B and C (since OA =OB = OC).

Since O is a unique point i.e. the point of intersection of perpendicular bisectors of AB and BC and we saw that OA=OB = OC , we can say that there is ONE and OnLY one circle that passes through all three non - collinear points.

Hence proved.