Proof of bpt theorem class 10

  1. Mid Point Theorem
  2. State and prove converse of BPT (Basic Proportionality Theorem).
  3. Basic Proportionality Theorem Class 10 Math Notes
  4. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
  5. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
  6. Basic Proportionality Theorem Class 10 Math Notes
  7. Mid Point Theorem
  8. State and prove converse of BPT (Basic Proportionality Theorem).


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Mid Point Theorem

Mid Point Theorem Geometry is one of the significant and essential branches of mathematics. This field deals with the geometrical problems and figures which are based on their properties. One of the important theorems in the field of geometry that deals with the properties of triangles are called the Mid- Point Theorem. Table of Contents: • • • • • • The theory of midpoint theorem is used in coordinate geometry, stating that the midpoint of the line segment is an average of the endpoints. The ‘x’ and the ‘y’ coordinates must be known for solving an equation using this theorem. The Mid- Point Theorem is also useful in the fields of calculus and algebra. Mid-Point Theorem Statement The midpoint theorem states that “ The line segment in a triangle joining the midpoint of any two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.” Mid-Point Theorem Proof If a line segment adjoins the mid-point of any two sides of a triangle, then the line segment is said to be parallel to the remaining third side and its measure will be half of the third side. Consider the triangle ABC, as shown in the above figure, Let E and D be the midpoints of the sides AC and AB. Then the line DE is said to be parallel to the side BC, whereas the side DE is half of the side BC; i.e., DE || BC DE = (1/2 *Â BC). Now consider the below figure, Construction-Â Extend the line segment DE and produce it to F such that, EF = DE and join CF. In triangle...

State and prove converse of BPT (Basic Proportionality Theorem).

Hint: Suppose another line that is parallel to the side (which is not getting intersected with the line which divides two sides in the same ratio) other than the line which divides the sides in the same ratio and basic proportionality theorem can be given as: If a line is drawn parallel to one side of a triangle and intersecting other two sides, then line divides the intersecting sides in the same ratio. Use this property to solve the given problem. Complete step by step answer: BPT stands for basic proportionality theorem. It states that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. So, if we draw a triangle ABC as Where $PQ||BC$ , so according to basic proportionality theorem as stated above we have, $\dfrac.................\left( i \right)$ Hence, converse of basic proportionality theorem can be stated as if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle. So, let us suppose a $\Delta ABC$ such that DE is a line which divides the two sides in the same ratio and draw a line DE’ which is parallel to the other side BC as shown in the diagram below: Here, DE is a line which divides AB and AC in the same ratio and DE is parallel to side BC. So, we have$\begin$ was parallel to the side BC (by construction). Hence, the converse of BPT is proved. Note: Another approach for solving this prop...

Basic Proportionality Theorem Class 10 Math Notes

Basic Proportionality Theorem (Thales Theorem) THEOREM 1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively. To prove: Construction: Join BE and CD and draw DM ⊥ AC and EN ⊥ AB. Proof: area of Δ ADE (Taking AD as base) So, [The area of Δ ADE is denoted as ar (ADE)]. Similarly, [Δ BDE and DEC are on the same base DE and between the same parallels BC and DE.] Therefore, from (i), (ii) and (iii), we have: . Corollary: From above equation we have . Adding '1' to both sides we have THEOREM 2: (Converse of BPT theorem) If a line divides any two sides of a triangle in the same ratio, prove that it is parallel to the third side. Given: In ΔABC, DE is a straight line such that . To prove: DE || BC. Construction: If DE is not parallel to BC, draw DF meeting AC at F. Proof: In ΔABC, let DF || BC [∴ A line drawn parallel to one side of a Δ divides the other two sides in the same ratio.] But . …(ii) [given] From (i) and (ii), we get . ⇒ FC = EC. It is possible only when E and F coincide Hence, DE || BC. SOME IMPORTANT RESULTS AND THEOREMS: (i) The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. (ii) In a triangle ABC, if D is a point on BC such that D divides BC in the ratio AB : AC,...

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Through a given point P ( 5 , 2 ) secants are drawn to cut the circle x 2 + y 2 = 2 5 at points A 1 ​ ( B 1 ​ ) , A 2 ​ ( B 2 ​ ) , A 3 ​ ( B 3 ​ ) , A 4 ​ ( B 4 ​ ) , A 5 ​ ( B 5 ​ ) such that P A 1 ​ + P B 1 ​ = 5 , P A 2 ​ + P B 2 ​ = 6 , P A 3 ​ + P B 3 ​ = 7 , P A 4 ​ + P B 4 ​ = 8 , P A 5 ​ + P B 5 ​ = 9. Find the value of ∑ i = 1 5 ​ P A i 2 ​ + ∑ i = 1 5 ​ P B i 2 ​ [Note: A r ​ ( B r ​ ) denotes that the line passing through P ( 5 , 2 ) meets the circle x 2 + y 2 = 2 5 at two points A r ​ and B r ​

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Through a given point P ( 5 , 2 ) secants are drawn to cut the circle x 2 + y 2 = 2 5 at points A 1 ​ ( B 1 ​ ) , A 2 ​ ( B 2 ​ ) , A 3 ​ ( B 3 ​ ) , A 4 ​ ( B 4 ​ ) , A 5 ​ ( B 5 ​ ) such that P A 1 ​ + P B 1 ​ = 5 , P A 2 ​ + P B 2 ​ = 6 , P A 3 ​ + P B 3 ​ = 7 , P A 4 ​ + P B 4 ​ = 8 , P A 5 ​ + P B 5 ​ = 9. Find the value of ∑ i = 1 5 ​ P A i 2 ​ + ∑ i = 1 5 ​ P B i 2 ​ [Note: A r ​ ( B r ​ ) denotes that the line passing through P ( 5 , 2 ) meets the circle x 2 + y 2 = 2 5 at two points A r ​ and B r ​

Basic Proportionality Theorem Class 10 Math Notes

Basic Proportionality Theorem (Thales Theorem) THEOREM 1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively. To prove: Construction: Join BE and CD and draw DM ⊥ AC and EN ⊥ AB. Proof: area of Δ ADE (Taking AD as base) So, [The area of Δ ADE is denoted as ar (ADE)]. Similarly, [Δ BDE and DEC are on the same base DE and between the same parallels BC and DE.] Therefore, from (i), (ii) and (iii), we have: . Corollary: From above equation we have . Adding '1' to both sides we have THEOREM 2: (Converse of BPT theorem) If a line divides any two sides of a triangle in the same ratio, prove that it is parallel to the third side. Given: In ΔABC, DE is a straight line such that . To prove: DE || BC. Construction: If DE is not parallel to BC, draw DF meeting AC at F. Proof: In ΔABC, let DF || BC [∴ A line drawn parallel to one side of a Δ divides the other two sides in the same ratio.] But . …(ii) [given] From (i) and (ii), we get . ⇒ FC = EC. It is possible only when E and F coincide Hence, DE || BC. SOME IMPORTANT RESULTS AND THEOREMS: (i) The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. (ii) In a triangle ABC, if D is a point on BC such that D divides BC in the ratio AB : AC,...

Mid Point Theorem

Mid Point Theorem Geometry is one of the significant and essential branches of mathematics. This field deals with the geometrical problems and figures which are based on their properties. One of the important theorems in the field of geometry that deals with the properties of triangles are called the Mid- Point Theorem. Table of Contents: • • • • • • The theory of midpoint theorem is used in coordinate geometry, stating that the midpoint of the line segment is an average of the endpoints. The ‘x’ and the ‘y’ coordinates must be known for solving an equation using this theorem. The Mid- Point Theorem is also useful in the fields of calculus and algebra. Mid-Point Theorem Statement The midpoint theorem states that “ The line segment in a triangle joining the midpoint of any two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.” Mid-Point Theorem Proof If a line segment adjoins the mid-point of any two sides of a triangle, then the line segment is said to be parallel to the remaining third side and its measure will be half of the third side. Consider the triangle ABC, as shown in the above figure, Let E and D be the midpoints of the sides AC and AB. Then the line DE is said to be parallel to the side BC, whereas the side DE is half of the side BC; i.e., DE || BC DE = (1/2 *Â BC). Now consider the below figure, Construction-Â Extend the line segment DE and produce it to F such that, EF = DE and join CF. In triangle...

State and prove converse of BPT (Basic Proportionality Theorem).

Hint: Suppose another line that is parallel to the side (which is not getting intersected with the line which divides two sides in the same ratio) other than the line which divides the sides in the same ratio and basic proportionality theorem can be given as: If a line is drawn parallel to one side of a triangle and intersecting other two sides, then line divides the intersecting sides in the same ratio. Use this property to solve the given problem. Complete step by step answer: BPT stands for basic proportionality theorem. It states that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. So, if we draw a triangle ABC as Where $PQ||BC$ , so according to basic proportionality theorem as stated above we have, $\dfrac.................\left( i \right)$ Hence, converse of basic proportionality theorem can be stated as if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle. So, let us suppose a $\Delta ABC$ such that DE is a line which divides the two sides in the same ratio and draw a line DE’ which is parallel to the other side BC as shown in the diagram below: Here, DE is a line which divides AB and AC in the same ratio and DE is parallel to side BC. So, we have$\begin$ was parallel to the side BC (by construction). Hence, the converse of BPT is proved. Note: Another approach for solving this prop...

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