Equal Intercept Theorem: In geometry, there are a lot of theorems. One of them is the equal intercept theorem. The theorem states that if a transversal forms equal intercepts on three or more parallel lines, any other line cutting them will also make equal intercepts. It means that given any three parallel lines, a line passing through them forms intercepts in the corresponding ratio of the distances between the lines. We use this theorem in other geometrical proofs such as midpoint theorem, converse midpoint theorem, etc. In this article, let us learn about the equal intercept theorem and its applications in detail.

Equal Intercept Theorem

If there are three or more parallel lines and the intercepts made by them on the transversal are equal, the corresponding intercepts on any other transversal are also equal.

Given: Lines \(l, m\) and \(n\) such that \(l||m||n,\) \(p\) is a transversal that cuts \(l, m, n in A, B, C\) respectively such that \(AB=BC.\) Also, \(q\) is another transversal.

Learn Basic Proportionality Theorem

To Prove: \(PQ=QR\)

Construction: Through \(Q,\) draw a line \(r\) such that \(r||p.\)

Proof:

Statement	Reason
\(r\parallel p\)	By construction Pair of opposite sides are parallel Opposite sides of a parallelogram are equal
\(ABQX\) is a parallelogram \(AB = XQ\,\,\,\,\,…\left( {\rm{i}} \right)\) \(BCQY\) is parallelogram \(BC = QY\,\,\,\,…\left( {{\rm{ii}}} \right)\) \(AB = BCXQ\, = \,\,QY\,\,\,…\left( {{\rm{iii}}} \right)\) Now, in \(\Delta PQX\) and \(\Delta RQY\) \(\angle PQX = \angle RQYXQ = QY\) \[\angle PQX = \angle RYQ\]	Pairs of opposite sides are parallel. Opposite sides of a parallelogram Given From (i) and (ii) Vertically opposite angles From Alternate angles, \(l\parallel n\) ASA congruency CPCT (iii)

Applications of Equal Intercept Theorem

We use the concept of equal intercept theorem in proving the basic proportionality theorem and mid-point theorem. Let us know about them in brief.

Basic Proportionality Theorem

The basic proportionality theorem was discovered 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, the line divides the two sides in the same ratio.

Let us consider a  \(\Delta ABC\), 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  \(\frac{A D}{D B}=\frac{A E}{E C}\).
Here, we can say that \(DE\) and \(BC\) are two parallel lines and \(AB, AC\) are transversal, which divides the parallel line in an equal ratio.
Hence, \(AD=AE\) and \(BD=CE\). So, \(\frac{A D}{D B}=\frac{A E}{E C}\)

Midpoint Theorem

The mid-point theorem states that if a line segment joins the mid-points of any two sides of a triangle, this line segment is parallel to the third side of the triangle and is half the length of the third side.

In \(\Delta ABC\) if \(D\) and \(E\) are mid-points of the sides \(AB\) and \(AC,\) respectively, then \(DE∥BC\) and \(D E=\frac{1}{2} B C\)

Converse of Midpoint Theorem

The converse of midpoint theorem states that if a straight line is drawn starting from the midpoint of any side of a triangle and parallel to another side, it intersects the third side at its mid-point.

In \(\Delta ABC\) if the line \(DE\) is drawn from the midpoint \(D\) of side \(AB\), parallel to \( BC\), it intersects the side third side \(AC,\) such that \(E\) is the midpoint of \(AC\) or \(AE=EC.\)

Example: Prove that if three parallel lines make equal intercepts on one transversal, then they will make equal intercepts on any other transversal using the midpoint theorem.

Let us first understand the problem better. Consider three lines and two transversals, as shown below:

Suppose that the intercepts on the left transversal are equal, that is, \(AB=BC.\)
We have to prove that the intercepts on the right transversal will also be equal, that is, \(DE=EF.\)

Construction: join \(A\) to \(F.\)

Let us consider \(\Delta ACF\)
Since \(B\) is the midpoint of \(AC\) and \(BG||CF,\) as \(BE||CF\), the converse of the midpoint theorem tells us that \(G\) is the midpoint of \(AF.\)

Now, consider \(\Delta AFD\), we have shown that \(G\) is the midpoint of \(AF\). Also, \(GE||AD.\)
Thus, the (converse of the) midpoint theorem tells us that \(E\) must be the midpoint of \(FD.\)
Hence, \(DE=EF.\)

Solved Examples – Equal Intercept Theorem

Q.1. In the triangle ABC, \(DE\parallel BC\), then find the value of EC?

Ans:
Given, \(DE∥BC.\)
According to the basic proportionality theorem, a line drawn parallel to one side divides the other sides in an equal ratio.
So, \(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\Rightarrow \frac{1.5}{3}=\frac{1}{E C}\)
\(\Rightarrow \frac{1}{2}=\frac{1}{E C}\)
\(\Rightarrow E C=2\)
Hence, the value of \(EC\) is \(2 \mathrm{~cm}\)

Q.2. Given, \(DE\parallel AC\) and \(DF\parallel AE\), prove that \(\frac{B F}{F E}=\frac{B E}{E C}\)

Ans:
In triangle \(ABC\), \(DE∥AC\)
According to the basic proportionality theorem, a line is drawn parallel to one side, dividing the remaining sides in an equal ratio.
\(\frac{B D}{D A}=\frac{B E}{E C}\) —– (1)
In triangle \(ABE,\) \(DF∥AE\)
According to the basic proportionality theorem, we know that a line drawn parallel to one side divides the remaining sides in an equal ratio.
\(\frac{B D}{D A}=\frac{B F}{F E}\)—– (2)
From (1) and (2),
\(\frac{B E}{F E}=\frac{B E}{E C}\)

Q.3. Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line, which joins the midpoints of the other sides of the triangle.
Ans: Given, In \(\Delta ABC\) \(E\) and \(F\) are midpoints of side \(AB\) and \(AC, \) respectively. \(AD\) is a line that meets \(BC\) at \(D.\) \(AD\) meets \(EF\) in \(R.\)
To Prove: \(AR=RD\)

Construction: Draw \(l||BC\) through \(A\)
Proof:

Statement	Reason
\(EF\parallel BC\,\,\,\,…\left( {\rm{i}} \right)\) \(l\parallel BC\,\,\,\,…\left( {{\rm{ii}}} \right)\) \(l\parallel EF\parallel BCAR\, = RD\)	Midpoint theorem, in \(\Delta ABC,E\) and \(F\) are midpoints of \(AB\) and \(AC\), respectively. Construction From (i) and (ii) Intercept theorem, \(l\parallel EF\parallel BC\) and \(AB\) and \(AD\) are transversals.

Q.4. In a  \(\Delta ABC\), \(D\) is the midpoint of BC. AC produced to E such that \(C E=\frac{1}{2} A C . E D\) produced meets AB at F and CP, DQ are drawn parallel to BA. Prove that \(DF=\frac{1}{3} E F\)
Ans: 
Given, In \(\Delta ABC,D\) is the midpoint of \(BC\)
\(P C|D Q| B A, C E=\frac{1}{2} A C, E D\) produced meets \(AB\) at \(F.\)
To prove: \(D F=\frac{1}{3} E F\)

Statement	Reason
\(A Q=Q C=\frac{1}{2} A C \ldots\)…(i) \(C E=\frac{1}{2} A C\)……(ii)	The converse of midpoint theorem. In \(D F=\frac{E F}{3}\) is the midpoint of \(BC\) and \(DQ||BA.\)
\(A Q=Q C=C E \ldots\) (iii) \(E P=P D=D F \ldots \ldots\) (iv)	Given From (i) and (ii) Intercept theorem,  \(EF\) and \(EA\) are transversals that intersect parallel lines
\(E F=3 D F\) \(D F=\frac{E F}{3}\)	\(PC, DQ, BA.\) From (iv) Simplification

Q.5. Consider a \(\Delta ABC\) and let D be any point on BC. Let X and Y be the midpoints of AB and AC. Show that XY will bisect AD.

Ans:
Given that \(X\) and \(Y\) are the midpoints of \(AB\) and \(AC\) in \(\Delta ABC\)
The midpoint theorem states that if a line segment joins the mid-points of any two sides of a triangle, this line segment is parallel to the third side of the triangle and is half the length of the third side.
Thus, \(XY||BC.\)
Now, consider \(\Delta ABD\)
The line segment \(XE\) is parallel to the base \(BD\) \((XY||BC)\)
According to the converse midpoint theorem, if a straight line is drawn starting from the midpoint of any side of a triangle and parallel to another side, it intersects the third side at its mid-point.
Thus, \(X\) is the midpoint of \(AB,\) and \(E\) must be the midpoint of \(AD.\)
Hence, \(XY\) bisects \(AD.\)

Summary

There are many theorems in geometry. The equal intercept theorem is one of them. The theorem states that if transversal forms equal intercepts on three or more parallel lines, any other line cutting them will also make equal intercepts. This article explained the proof and the applications of equal intercept theorem such as midpoint theorem, converse midpoint theorem, basic proportionality theorem etc. These are the extensions of the equal intercept theorem. This article stated some proofs using this theorem.

Learn About Midpoint Theorem

Frequently Asked Questions (FAQs)

Q.1. What is the equal intercept theorem?
Ans: If there are three or more parallel lines and the intercepts made by them on the transversal are equal, the corresponding intercepts on any other transversal are also equal.

Q.2. Write two applications of the equal intercept theorem
Ans: Two applications of equal intercept theorem are the basic proportionality theorem and converse midpoint theorem.

Q.3. What is the converse midpoint theorem?
Ans: The converse of midpoint theorem states that if a straight line is drawn starting from the midpoint of any side of a triangle and parallel to another side, it intersects the third side at its mid-point.

Q.4. Are intercepts made by parallel lines only?
Ans: Intercepts can be made by any line, but equal intercepts are made by parallel lines only.

Q.5. What is the basic proportionality theorem?
Ans: 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, the line divides the two sides in the same ratio.

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