BPT (basic proportionality theorem) Class 10
Hello dear friends, we will study about BPT (basic proportionality theorem) Class 10, including its applications and solved examples.
Basic Proportionality Theorem was given by a famous famous Greek Mathematician, Thales, hence it is also called Thales Theorem. According to him, for any two equiangular triangles, the ratio of any two corresponding sides is always the same. Based on this concept, he gave theorem of basic proportionality (BPT).
The BPT (basic proportionality theorem) Class 10 is crucial for understanding geometry and helps in solving various mathematical problems.
This concept has been introduced in similar triangles. If two triangles are similar to each other then,
i) Corresponding angles of both the triangles are equal.
ii) Corresponding sides of both the triangles are in proportion to each other.
In BPT (basic proportionality theorem) Class 10, we explore the relationships between the sides of similar triangles.
Basic Proportionality Theorem (Theorem 6.1)
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.
Proof : We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at P and Q respectively.
Next, we will apply the BPT (basic proportionality theorem) Class 10 in solving real-world problems.
Mastering the BPT (basic proportionality theorem) Class 10 can assist students in higher-level mathematics.
According to the basic proportionality theorem as stated above, we need to prove:
According to the basic proportionality theorem as stated above, we need to prove:
According to the basic proportionality theorem as stated above, we need to prove:
Understanding BPT (basic proportionality theorem) Class 10 can significantly enhance your problem-solving skills in mathematics.
Construction: (Basic Proportionality Theorem )
This theorem, known as BPT (basic proportionality theorem) Class 10, is a fundamental aspect of triangle geometry.
To summarize, the BPT (basic proportionality theorem) Class 10 provides essential insights into the properties of triangles.
Basic Proportionality Theorem (Theorem 6.1): 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. Proof: Given a triangle ABC, where a line parallel to side BC intersects sides AB and AC at points P and Q respectively.
Join the vertex B of ΔABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM⊥AC as shown in the given figure
Here, we will focus on the BPT (basic proportionality theorem) Class 10 and its various applications.
Proof
Now the area of ∆APQ = 1/2 × AP × QN (Since, area of a triangle= 1/2× Base × Height)
Similarly, area of ∆PBQ= 1/2 × PB × QN
area of ∆APQ = 1/2 × AQ × PM
Also,area of ∆QCP = 1/2 × QC × PM ………… (1)
Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have
Similarly
According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.
Therefore, we can say that ∆PBQ and QCP have the same area.
area of ∆PBQ = area of ∆QCP …………..(3)
Therefore, from the equations (1), (2) and (3) we can say that,
Also, ∆ABC and ∆APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that ∆ABC ~∆APQ.
The MidPoint theorem is a special case of the basic proportionality theorem.
According to mid-point theorem, a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.
Converse of B.P.T.T Theorem: (Theorem 6.2 )
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Proof :(Converse of B.P.T.T Theorem)
Suppose a line DE, intersects the two sides of a triangle AB and AC at D and E, such that;
AD/DB = AE/EC ……(1)
Assume DE is not parallel to BC. Now, draw a line DE’ parallel to BC.
Hence, by similar triangles,
AD/DB = AE’/E’C ……(2)
From eq. 1 and 2, we get;
AE/EC = AE’/E’C
Adding 1 on both the sides;
We encourage students to practice problems related to the BPT (basic proportionality theorem) Class 10 regularly.
Additionally, exploring the BPT (basic proportionality theorem) Class 10 helps in understanding advanced geometry concepts.
AE/EC + 1 = AE’/E’C +1
To conclude, the BPT (basic proportionality theorem) Class 10 is vital for a strong foundation in geometry.
(AE +EC)/EC = (AE’+E’C)/E’C
AC/EC = AC/E’C
So, EC = E’C
This is possible only when E and E’ coincide.
But, DE’ || BC
Therefore, DE ||BC.
Hence, proved.
Question (1) Example : If a line intersects sides AB and AC of a ∆ ABC at D and E respectively
and is parallel to BC, prove that
Solution:- DE || BC
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