Bpt prove

(i) AR/RD = AQ/AB Solution : PQ|| BC AP/PC = AQ/QB---(1) PR||CD AP/PC = AR/RD ---(2) (1) = (2) AQ/QB = AR/RD (ii) QB/AQ = DR/AR Solution : Taking reciprocal on the first result, we get QB/AQ = DR/AR Question 2 : Rhombus PQRB is inscribed in Δ ABC such that ÐB is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus Solution : In triangle ABC, BC is parallel to PQ. Let side of rhombus be a cm. Given AB = 12 cm. So AP = 12 – a BC = 6 cm We have AP / AB = PQ / BC 12 – a / 12 = a / 6 12 a = 72 – 6 a 18 a = 72 a = 72 / 18 a = 4 cm PQ = 4 cm and RB = 4 cm Question 3 : In trapezium ABCD, AB ||DC , E and F are points on non-parallel sides AD and BC respectively, such that EF ||AB . Show that AE/ED = BF/FC Solution : In triangle DAB, DE/EA = DT/TB -----(1) In triangle DBC BT/TD =BF/FC -----(2) Now let us take the reciprocal of (2) and equate them with (1). So we get TD/BT = FC/BF DE/DA = FC/BF Take reciprocal on both sides, we get DA/DE = BF/FC Hence proved. Question 4 : In figure DE || BC and CD|| EF. Prove that AD 2 = AB×AF After having gone through the stuff given above, we hope that the students would have understood, " Practice Problems Based on bpt Theorem". Apart from the stuff given in this section " Practice Problems Based on bpt Theorem" , if you need any other stuff in math, please use our google custom search here.

I've got exercise to do as en exercise to my school leaving exam and I have no idea how to prove it: Diagonals of trapezium intersect in point $S$. Through point $S$ the segment was given that is parallel to the bases and intersect the legs in points $E$ and $F$. Prove that $|ES| = |SF|$. I'm sorry for my poor translation but I'm not good at math english yet. Sol: Given: ABCD is a trapezium. To Prove: DE/EB = CE /EA Construction: Draw EF || BA || CD, intersecting AD in F. Proof: FE || AB (Given) DE/EB = DF/FA [According to basic proportionality theorem] --------- (1) FE || DC (Given) CE/EA = DF/FA [According to basic proportionality theorem] --------- (2) From (1) and (2), we get DE/EB = CE/EA So, diagonals of a trapezium divide each other proportionally.

The converse of the Basic proportionality theorem (BPT): According to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. The Basic Proportionality Theorem, often known as the Thales Theorem, was developed by the eminent Greek mathematician Thales. He asserted that the ratio of any two matching sides is constant for any two equiangular triangles. He developed the basic proportionality theorem based on this idea (BPT). Prove Converse of BPT Let's say that a line DE intersects AB and AC's two sides at D and E, as shown; AD/DB = AE/EC ….. (i) Assume DE and BC are not parallel. Next, connect BC and DE' with a straight line. thus, using similar triangles: AD/DB = AE’/E’C …. (ii) From eq. (i) and (ii), we get; AE/EC = AE’/E’C Adding 1 on both sides: AE/EC + 1 = AE’/E’C + 1 (AE + EC)/ EC = (AE’ + E’C)/ E’C AC/EC = AC/E’C EC = E’C This is only feasible if E and E' coincide. But DE’ || BC Therefore, DE || BC Hence, proved Summary: State and prove converse of BPT (basic proportionality theorem). Contrary to the Basic Proportionality Theorem (BPT), a line is parallel to the third side if it divides any two sides of a triangle in the same ratio. It's been proven.

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The Basic Proportionality Theorem was developed by “Thales,” a prominent Greek mathematician. The Thales theorem is another name for this theory. “A line drawn parallel to one side of a triangle and cutting the other two sides splits the other two sides in equal proportion,” says the basic proportionality theorem, often known as the Thales theorem. The lengths of the intercepts made by a line segment on the other two sides of a triangle when it is drawn parallel to the third side of the triangle are described by this theorem. It is an important theorem in geometry that deals with triangles. We recommend that students pay great attention to this theorem since it will help them in understanding basic mathematical concepts such as triangles in geometry. The more they understand, the easier it will be for them to achieve excellent exam scores. Definition of Basic Proportionality Theorem The basic proportionality theorem was found by a famous Greek mathematician Thales. It is also called the Thales theorem. The Basic proportionality Theorem or Thales Theorem states that if a line is drawn parallel to one side of a triangle and intersect the other two sides, then the line divides the two sides in the same ratio. Let us consider a triangle \(ABC,\) such that line \(DE\) has drawn parallel to the base of the triangle \(BC.\) Then, the line \(DE\) divides the sides \(AB, AC\) in the same ratio.So, from the Basic proportionality Theorem or Thales Theorem, we get \(\frac\) Proof of B...

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The converse of the Basic proportionality theorem (BPT): According to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. The Basic Proportionality Theorem, often known as the Thales Theorem, was developed by the eminent Greek mathematician Thales. He asserted that the ratio of any two matching sides is constant for any two equiangular triangles. He developed the basic proportionality theorem based on this idea (BPT). Prove Converse of BPT Let's say that a line DE intersects AB and AC's two sides at D and E, as shown; AD/DB = AE/EC ….. (i) Assume DE and BC are not parallel. Next, connect BC and DE' with a straight line. thus, using similar triangles: AD/DB = AE’/E’C …. (ii) From eq. (i) and (ii), we get; AE/EC = AE’/E’C Adding 1 on both sides: AE/EC + 1 = AE’/E’C + 1 (AE + EC)/ EC = (AE’ + E’C)/ E’C AC/EC = AC/E’C EC = E’C This is only feasible if E and E' coincide. But DE’ || BC Therefore, DE || BC Hence, proved Summary: State and prove converse of BPT (basic proportionality theorem). Contrary to the Basic Proportionality Theorem (BPT), a line is parallel to the third side if it divides any two sides of a triangle in the same ratio. It's been proven.

I've got exercise to do as en exercise to my school leaving exam and I have no idea how to prove it: Diagonals of trapezium intersect in point $S$. Through point $S$ the segment was given that is parallel to the bases and intersect the legs in points $E$ and $F$. Prove that $|ES| = |SF|$. I'm sorry for my poor translation but I'm not good at math english yet. Sol: Given: ABCD is a trapezium. To Prove: DE/EB = CE /EA Construction: Draw EF || BA || CD, intersecting AD in F. Proof: FE || AB (Given) DE/EB = DF/FA [According to basic proportionality theorem] --------- (1) FE || DC (Given) CE/EA = DF/FA [According to basic proportionality theorem] --------- (2) From (1) and (2), we get DE/EB = CE/EA So, diagonals of a trapezium divide each other proportionally.

(i) AR/RD = AQ/AB Solution : PQ|| BC AP/PC = AQ/QB---(1) PR||CD AP/PC = AR/RD ---(2) (1) = (2) AQ/QB = AR/RD (ii) QB/AQ = DR/AR Solution : Taking reciprocal on the first result, we get QB/AQ = DR/AR Question 2 : Rhombus PQRB is inscribed in Δ ABC such that ÐB is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus Solution : In triangle ABC, BC is parallel to PQ. Let side of rhombus be a cm. Given AB = 12 cm. So AP = 12 – a BC = 6 cm We have AP / AB = PQ / BC 12 – a / 12 = a / 6 12 a = 72 – 6 a 18 a = 72 a = 72 / 18 a = 4 cm PQ = 4 cm and RB = 4 cm Question 3 : In trapezium ABCD, AB ||DC , E and F are points on non-parallel sides AD and BC respectively, such that EF ||AB . Show that AE/ED = BF/FC Solution : In triangle DAB, DE/EA = DT/TB -----(1) In triangle DBC BT/TD =BF/FC -----(2) Now let us take the reciprocal of (2) and equate them with (1). So we get TD/BT = FC/BF DE/DA = FC/BF Take reciprocal on both sides, we get DA/DE = BF/FC Hence proved. Question 4 : In figure DE || BC and CD|| EF. Prove that AD 2 = AB×AF After having gone through the stuff given above, we hope that the students would have understood, " Practice Problems Based on bpt Theorem". Apart from the stuff given in this section " Practice Problems Based on bpt Theorem" , if you need any other stuff in math, please use our google custom search here.

The Basic Proportionality Theorem was developed by “Thales,” a prominent Greek mathematician. The Thales theorem is another name for this theory. “A line drawn parallel to one side of a triangle and cutting the other two sides splits the other two sides in equal proportion,” says the basic proportionality theorem, often known as the Thales theorem. The lengths of the intercepts made by a line segment on the other two sides of a triangle when it is drawn parallel to the third side of the triangle are described by this theorem. It is an important theorem in geometry that deals with triangles. We recommend that students pay great attention to this theorem since it will help them in understanding basic mathematical concepts such as triangles in geometry. The more they understand, the easier it will be for them to achieve excellent exam scores. Definition of Basic Proportionality Theorem The basic proportionality theorem was found by a famous Greek mathematician Thales. It is also called the Thales theorem. The Basic proportionality Theorem or Thales Theorem states that if a line is drawn parallel to one side of a triangle and intersect the other two sides, then the line divides the two sides in the same ratio. Let us consider a triangle \(ABC,\) such that line \(DE\) has drawn parallel to the base of the triangle \(BC.\) Then, the line \(DE\) divides the sides \(AB, AC\) in the same ratio.So, from the Basic proportionality Theorem or Thales Theorem, we get \(\frac\) Proof of B...