Mid-Point Theorem and its converse, equal intercept theorem | ICSE Class 9 Maths Notes

This note covers midpoints, the mid-point theorem and its proof, the converse and its proof, equal intercepts made by parallel lines, and applications to triangles, quadrilaterals and bisected segments.

What do the terms and symbols in these theorems mean?

How are points, segments and midpoints named?

A line segment is the part of a straight line between two endpoints. Capital letters name points. For endpoints A and B, AB names their segment or its length, as the context requires. A line through A and B extends beyond both endpoints. Points on one straight line are called collinear.

A midpoint divides a segment into two equal parts. Thus, if P is the midpoint of AB, P lies on AB and AP = PB. The symbol = means “is equal to”. Writing AP = AB/2 means that AP is half the length of AB.

Definition: To bisect a segment means to divide it into two equal segments. A statement that a line bisects AB therefore identifies an intersection point on AB with equal distances from A and B.

A triangle has three sides and three vertices, or corner points. The notation triangle ABC names the triangle with vertices A, B and C. Its sides are AB, BC and CA. The side not containing either of two chosen midpoints is the third side.

What do parallelism and congruence mean?

Parallel lines lie in the same plane, a flat surface extending in all directions, and do not meet however far extended. The symbol ∥ means “is parallel to”. Segments are called parallel when the lines containing them are parallel. A transversal is a line cutting two or more lines at distinct points.

Congruent triangles have the same shape and size, with equal corresponding sides and angles. Corresponding parts are those matched in the stated order of vertices. The notation ∠APQ means the angle at P formed by PA and PQ; the middle letter names the vertex.

A theorem is a statement established by proof. Its hypotheses are the conditions given, and its conclusion is what follows. An auxiliary construction adds a useful line or segment to a figure. Such a construction must be stated before its properties are used.

What does the mid-point theorem state, and how is it proved?

Theorem: Mid-point theorem

The segment joining the midpoints of two sides of a triangle is parallel to the third side and has half its length. There are two conclusions: one concerns direction, and the other concerns length. A complete statement includes both.

In triangle ABC, let P and Q be the midpoints of AB and AC respectively, meaning in that order. Then AP = PB and AQ = QC. The theorem gives PQ ∥ BC and PQ = BC/2. Here P and Q lie on different sides meeting at A.

Triangle ABC contains P on AB and Q on AC. The line through P and Q continues to R. A line named l passes through C and R, parallel to AB. Matching marks show the midpoint divisions.

See Fig. 12.14 in your NCERT textbook

How does the auxiliary line help?

Draw a line through C parallel to BA, meeting the extended line PQ at R. The letter l is simply the name of this new line. Compare triangles APQ and CRQ. The construction supplies parallel sides, while the given midpoint supplies an equal side.

Vertically opposite angles are the opposite angles formed when two straight lines intersect; they are equal. Alternate interior angles lie between parallel lines on opposite sides of a transversal; they are equal. These angle facts connect the two triangles in the construction.

1. AQ = CQ because Q is the midpoint of AC. Also, ∠AQP = ∠CQR because these angles are vertically opposite.
2. Since AP ∥ CR and P, Q, R lie on one straight line, ∠APQ = ∠CRQ by the alternate interior angle property.
3. Triangles APQ and CRQ are congruent by AAS, the angle-angle-side test: two corresponding angles and a corresponding non-included side are equal. The non-included side is not the side joining the two specified angle vertices.
4. Corresponding sides give PQ = QR and AP = CR. Since AP = PB, it follows that PB = CR.
5. A parallelogram is a quadrilateral with both pairs of opposite sides parallel. A quadrilateral is a four-sided plane figure. The equal and parallel opposite sides PB and CR make BCRP a parallelogram.
6. Opposite sides of a parallelogram are equal and parallel. Therefore PR ∥ BC and PR = BC. Since PQ = QR, PQ = PR/2 = BC/2, and PQ ∥ BC.

Worked example 1. In triangle ABC, P and Q are the midpoints of AB and AC. Express the length PQ as a fraction of BC.

Answer: Both midpoint conditions hold, so the mid-point theorem gives PQ = (1/2)BC. The fraction is one-half. The same theorem also establishes PQ ∥ BC; the length conclusion does not replace the parallelism conclusion.

What is the converse of the mid-point theorem?

Theorem: Converse of the mid-point theorem

A line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side. The resulting segment also has half the length of the parallel side. The known midpoint and the given parallelism are both essential conditions.

In triangle ABC, let P be the midpoint of AB. Draw PQ parallel to BC, meeting AC at Q. The conclusion is AQ = QC, so Q is the midpoint of AC. Having obtained both midpoints, we also have PQ = BC/2.

P lies on AB, and the horizontal line through P meets AC at Q and continues to R. The line CR is parallel to AB. Matching arrows mark PR and BC as parallel.

See Fig. 12.15 in your NCERT textbook

How can the parallelogram proof be reversed?

As before, draw CR parallel to BA to meet the line PQ produced at R. Here “produced” means extended beyond its endpoint. Now PR ∥ BC is already known, so BCRP is a parallelogram immediately. Its opposite sides yield CR = BP.

1. Because P is the midpoint of AB, BP = PA. Combining this with CR = BP gives CR = PA.
2. Since AP ∥ CR, ∠APQ = ∠CRQ. Also, ∠AQP = ∠CQR because they are vertically opposite angles.
3. Triangles APQ and CRQ are congruent by AAS. Corresponding sides therefore give AQ = CQ and PQ = QR.
4. Thus Q bisects AC. Since PR = BC and PQ = QR, the additional length conclusion is PQ = PR/2 = BC/2.

Why is there also a shorter proof?

Let M be the midpoint of AC. By the mid-point theorem, PM ∥ BC. But the given line PQ also passes through P and is parallel to BC. There is a unique parallel, meaning exactly one parallel line, to a given line through a point outside that line.

Consequently PM and PQ are the same line. They meet AC at the same point, so M and Q coincide, meaning they are the same point. Since M is the midpoint, Q is the midpoint too. The uniqueness of the parallel is the key additional reason.

Worked example 2. In triangle ABC, P is the midpoint of AB and PQ ∥ BC, with Q on AC. Express AQ and QC in terms of AC.

Answer: The converse gives AQ = QC. Their sum is AC, so AQ = QC = (1/2)AC. After establishing Q as a midpoint, the direct theorem also gives PQ = (1/2)BC.

How do you distinguish the theorem from its converse?

Which facts are given, and which must be proved?

A converse reverses the direction of an implication, an “if ... then ...” statement, while retaining the relevant setting. Here the useful comparison keeps P as a known midpoint. The direct theorem uses Q as another midpoint to establish parallelism; the converse uses parallelism to establish that Q is a midpoint.

Read “respectively” carefully: it pairs items in their stated order. If P and Q are midpoints of AB and AC respectively, P belongs to AB and Q belongs to AC. The pair of sides containing the midpoints determines the third side; exchanging the labels P and Q between AB and AC leaves PQ and its relationship to BC unchanged.

What makes a valid chain of reasoning?

Write the triangle name before applying a result. Then identify its two relevant sides and the segment under discussion. A diagram can contain several triangles, and a line that is a side of one triangle may pass through the interior of another.

Keep each reason next to the conclusion it establishes. For example, “P and Q are midpoints, so PQ ∥ BC” invokes the theorem. “P is a midpoint and PQ ∥ BC, so AQ = QC” invokes the converse. Neither argument should assume its own conclusion.

The truth of a theorem does not by itself prove its converse. The shorter converse proof works because it adds the uniqueness of the parallel through P. This is why the extra reasoning matters even though the direct theorem appears inside the proof.

How do three midpoints divide a triangle into four congruent triangles?

What lengths follow from applying the theorem three times?

Let P, Q and R be the midpoints of AB, AC and BC respectively in triangle ABC. Join PQ, QR and RP. Use each pair of midpoints separately: the third side changes with the chosen pair.

Triangle ABC has P on AB, Q on AC and R on BC. Segments PQ, QR and RP divide it into a central triangle and three corner triangles. Matching side marks indicate the midpoint divisions.

See Fig. 12.22 in your NCERT textbook

Worked example 3. P, Q and R are the midpoints of AB, AC and BC in triangle ABC. Find PQ, QR and PR in terms of the original sides.

Answer: Applying the mid-point theorem to each pair gives PQ = (1/2)BC, QR = (1/2)AB and PR = (1/2)AC. Each joining segment is parallel to the side whose half-length appears in its expression.

How do equal sides establish congruence?

SSS means side-side-side: triangles are congruent if their three corresponding sides are equal. Each small triangle has one side of length AB/2, one of length AC/2 and one of length BC/2. The order of matching vertices must preserve these correspondences.

Worked example 4. With the same three midpoint conditions in triangle ABC, prove that triangle PQR is congruent to the three corner triangles.

Answer: The 4 triangles have the same three side lengths. For the first comparison, PQ = QP, QR = PA and PR = QA. Thus triangle PQR is congruent to triangle QPA by SSS. The other correctly ordered comparisons are triangle PQR with triangle RBP and with triangle CRQ.

For triangle RBP, PQ = RB, QR = BP and PR = RP. For triangle CRQ, PQ = CR, QR = RQ and PR = CQ. These equalities either come from the midpoint definitions or from the theorem, and complete the remaining SSS comparisons.

The congruence conclusion concerns the four smaller triangles. None is claimed to be congruent to the original triangle ABC. Their side lengths are half the corresponding original lengths, whereas congruence requires equal corresponding lengths.

How can a midpoint line bisect another segment inside a triangle?

Which smaller triangle should you use?

In triangle ABC, let M and N be the midpoints of AB and AC. Let D be a point on BC. The segment AD joins A to that point. The aim is to prove that MN bisects AD, even though D need not be the midpoint of BC.

M and N lie on AB and AC, with matching marks showing equal halves. D lies on BC between B and C. The segment AD crosses MN inside triangle ABC.

See Fig. 12.23 in your NCERT textbook

First apply the direct theorem in the larger triangle ABC. It gives MN ∥ BC. Now let X name the intersection of MN with AD. Because D lies on BC, the segment BD lies along the same straight line as BC.

Worked example 5. In triangle ABC, M and N are midpoints of AB and AC. D lies on BC, and X is the intersection of AD with MN. Prove that AX = XD.

Answer: MN ∥ BC by the mid-point theorem. If D is different from B and C, use triangle ABD: M is the midpoint of AB and MX ∥ BD. The converse gives AX = XD = (1/2)AD.

Why does the location of D matter in the proof?

The construction introduces X only to name the point whose midpoint property is being proved. It does not assume that X is a midpoint. Parallelism comes first; then the converse supplies the equality AX = XD.

If D is B, AD is AB and its intersection with MN is the known midpoint M. If D is C, the corresponding point is N on AC. These endpoint cases follow directly from the given midpoint conditions and do not require a smaller triangle.

This application links the two results in sequence. The first theorem creates a useful parallel line, and the converse transfers a known midpoint to another segment. State the triangle used at each stage to make that transfer explicit.

How do the midpoints of a quadrilateral form a parallelogram?

How do diagonals reveal the necessary triangles?

Let ABCD be a convex quadrilateral: its interior angles, the angles inside the figure, are each less than a straight angle. A straight angle is a half-turn. A diagonal joins non-adjacent vertices, meaning vertices not joined by a side. Its diagonals AC and BD meet inside it.

Let P, Q, R and S be the midpoints of AB, BC, CD and DA respectively. Join them in that order. Drawing the diagonals makes triangles available for repeated applications of the mid-point theorem.

ABCD surrounds the quadrilateral PQRS. P, Q, R and S are marked on successive sides. The diagonals AC and BD appear as dashed lines, and matching marks indicate equal halves of the outer sides.

See Fig. 12.21 in your NCERT textbook

Worked example 6. In convex quadrilateral ABCD, P, Q, R and S are the midpoints of AB, BC, CD and DA. Prove that PQRS is a parallelogram.

Answer: In triangles ABC and ADC, the theorem gives PQ ∥ AC, SR ∥ AC and PQ = SR = (1/2)AC. Thus PQ ∥ SR. In triangles BCD and BAD, QR ∥ BD and PS ∥ BD, so QR ∥ PS. Both pairs of opposite sides are parallel.

What follows about the new diagonals?

The midpoint quadrilateral is called the Varignon parallelogram. Its sides are related to the diagonals of the original quadrilateral. In particular, PQ and SR each equal AC/2, while QR and PS each equal BD/2.

Worked example 7. For the same convex quadrilateral and four specified midpoints, let O be the intersection of PR and QS. Show that PR and QS bisect each other.

Answer: The midpoint theorem makes PQRS a parallelogram, as established above. PR and QS are its diagonals. The diagonals of a parallelogram bisect each other, so OP = OR = (1/2)PR and OQ = OS = (1/2)QS.

Keep the two figures distinct. AC and BD belong to the original quadrilateral; PR and QS belong to the midpoint parallelogram. The proof that PR and QS bisect each other uses the newly established parallelogram, not an unstated property of ABCD.

The conclusion that PQRS is a parallelogram does not require ABCD itself to be a parallelogram. The required starting information consists of the four midpoint conditions. The diagonals are auxiliary segments that expose the triangles in which those conditions can be used.

What is the equal intercept theorem, and how is it proved?

Theorem: Equal intercept theorem

If three or more parallel lines make equal intercepts on one transversal, they make equal intercepts on any other transversal cutting them. An intercept here is the segment of a transversal between two of the parallel lines. Consecutive intercepts lie between successive parallel lines.

For three parallel lines, let one transversal meet them at A, B and C in order. Let another meet them at D, E and F in the corresponding order. Take these as six distinct points. Thus AD ∥ BE ∥ CF. If AB = BC, then DE = EF.

Draw three distinct parallel lines. Mark A, B, C successively on a transversal and D, E, F successively on another, with A and D on the first parallel. Mark AB = BC. Join AF, meeting the middle parallel at X.

How does the converse of the mid-point theorem prove it?

The segment AF is an auxiliary diagonal joining a point on the first parallel to a point on the third. Its intersection X with the middle parallel creates two triangles in which the converse can be applied successively.

1. In triangle ACF, B lies on AC and AB = BC. Therefore B is the midpoint of AC.
2. The line BX lies on the middle parallel, so BX ∥ CF. By the converse of the mid-point theorem, X is the midpoint of AF. Hence AX = XF.
3. Now consider triangle FAD. X is the midpoint of FA, and XE ∥ AD because the middle and first lines are parallel.
4. Applying the converse in triangle FAD shows that E is the midpoint of FD. Therefore DE = EF, which is the required equality of intercepts.

The proof transfers a midpoint first from AC to AF and then from AF to DF. It uses two applications of the converse, not a measurement of the drawing. Both triangle names and both parallel relationships are needed.

For more than three parallel lines, apply the three-line result to successive groups of three. Each group gives equality of the corresponding adjacent intercepts on the second transversal. Linking those equalities shows that all the consecutive intercepts there are equal.

How should equal intercepts be applied and checked?

Which lengths are equal?

Return to the arrangement AD ∥ BE ∥ CF, with A, B, C in order on one transversal and D, E, F in corresponding order on the other. Equal intercepts AB and BC lead to equal intercepts DE and EF.

The conclusion compares lengths on the same transversal. It does not assert that AB equals DE. The second transversal can cut the parallels at a different angle, so equality within each sequence must not be confused with equality between the two sequences.

Worked example 8. Three parallel lines meet one transversal at A, B, C in order and another at D, E, F in corresponding order. Given AB = BC, express DE and EF in terms of DF.

Answer: By the equal intercept theorem, DE = EF. Since E lies between D and F, DE + EF = DF. Therefore DE = EF = (1/2)DF. No numerical length is needed to establish either equality.

How can the conditions be recorded clearly?

Check the whole segment before halving it. In this arrangement DF consists of DE followed by EF, whereas AC consists of AB followed by BC. Halving DF gives its own intercepts; it does not give lengths on the other transversal.

The theorem can also be applied with the roles of the transversals exchanged. If the same parallels give DE = EF, they give AB = BC. This uses the stated theorem with a different starting transversal; it does not remove the parallel-line condition.

Across all three theorems, organise a solution around the available information: two midpoints, a midpoint with a parallel, or equal intercepts with parallel lines. Then name the conclusion precisely as parallelism, half-length, or bisection, and justify each additional deduction separately.

Glossary

• Midpoint — A point on a segment that divides it into two equal lengths.
• Bisection — Division of a line segment into two equal parts at its midpoint.
• Parallel lines — Lines in the same plane that do not meet however far extended.
• Transversal — A line that cuts two or more other lines at distinct points.
• Intercept — The segment cut off on a transversal between two specified parallel lines.
• Converse — A statement reversing an implication while keeping its relevant setting and fixed conditions.
• Congruent triangles — Triangles with the same shape and size and equal corresponding sides and angles.
• Corresponding parts — Sides or angles matched by the stated order of vertices in compared triangles.
• Auxiliary construction — An added line or segment that supplies relationships useful in a geometric proof.
• Parallelogram — A quadrilateral in which both pairs of opposite sides are parallel.
• Diagonal — A segment joining two vertices of a quadrilateral that are not adjacent.
• Vertically opposite angles — Opposite angles formed by two intersecting straight lines, equal in measure.

Common errors and misconceptions

• Misconception: One midpoint is enough to use the direct mid-point theorem. Correct: The joining segment must connect the midpoints of two sides of the same triangle.
• Misconception: The mid-point theorem gives only a parallel line. Correct: It also states that the joining segment has half the third side's length.
• Misconception: A line through a midpoint bisects another side without any further condition. Correct: For the converse, that line must also be parallel to the appropriate third side.
• Misconception: The four small triangles formed by joining midpoints are congruent to the original triangle. Correct: They are congruent to one another, with side lengths half the corresponding original lengths.
• Misconception: The midpoint quadrilateral is a parallelogram only when the original quadrilateral is one. Correct: Applying the theorem in triangles formed by the diagonals establishes the midpoint parallelogram.
• Misconception: Equal intercepts mean equal lengths across different transversals. Correct: Equality is transferred between consecutive intercepts within each transversal; the theorem does not equate the two transversals' intercept lengths.
• Misconception: A segment that looks bisected in a diagram needs no proof. Correct: Use stated midpoint information, congruence or an applicable theorem to justify equality of the two parts.

Exam-style questions with model answers

Q1. In triangle ABC, P and Q are the midpoints of AB and AC respectively. State the two conclusions of the mid-point theorem for PQ. [2 marks]

1. PQ ∥ BC: the segment joining the two given midpoints is parallel to the third side of triangle ABC.
2. PQ = BC/2: the same joining segment has half the length of that third side.

Q2. In triangle ABC, P is the midpoint of AB. The line through P parallel to BC meets AC at Q. Prove that Q is the midpoint of AC by considering the midpoint M of AC. [3 marks]

1. Let M be the midpoint of AC. Since P and M are midpoints of two sides of triangle ABC, the mid-point theorem gives PM ∥ BC.
2. Both PM and PQ pass through P and are parallel to BC. The uniqueness of the parallel through P means they are the same line.
3. That line meets AC at one point, so M and Q coincide. Since M is the midpoint of AC, AQ = QC, proving that Q is its midpoint.

Q3. In triangle ABC, P and Q are midpoints of AB and AC. Draw CR parallel to BA, meeting the extended line PQ at R. Prove PQ ∥ BC and PQ = BC/2. [5 marks]

1. AQ = CQ because Q is the midpoint of AC. Also, ∠AQP = ∠CQR since the straight lines AC and PR form vertically opposite angles at Q.
2. AP ∥ CR by construction, so ∠APQ = ∠CRQ by alternate interior angles. Thus triangles APQ and CRQ are congruent by AAS.
3. Corresponding sides give PQ = QR and AP = CR. The other midpoint condition is AP = PB, so PB = CR.
4. PB and CR are equal and parallel opposite sides of BCRP. Hence BCRP is a parallelogram, whose opposite sides give PR ∥ BC and PR = BC.
5. Since P, Q and R are collinear, PQ ∥ BC. Also PQ = QR, so PQ = PR/2 = BC/2, establishing both required conclusions.

Q4. In triangle ABC, M and N are midpoints of AB and AC. D lies strictly between B and C, and AD meets MN at X. Prove AX = XD. [3 marks]

1. In triangle ABC, M and N are the midpoints of two sides. The mid-point theorem therefore gives MN ∥ BC.
2. Since X lies on MN and D lies on BC, MX ∥ BD. Now consider triangle ABD, in which M is the midpoint of AB.
3. In that triangle, MX passes through the midpoint M and is parallel to BD. By the converse of the mid-point theorem, it bisects AD, so AX = XD.

Q5. In convex quadrilateral ABCD, P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Prove that PQRS is a parallelogram by using diagonals AC and BD. [4 marks]

1. Draw diagonal AC. In triangle ABC, P and Q are midpoints, so the mid-point theorem gives PQ ∥ AC.
2. In triangle ADC, S and R are midpoints, so SR ∥ AC. Thus PQ ∥ SR because both are parallel to AC.
3. Draw diagonal BD. In triangles BCD and BAD, the mid-point theorem gives QR ∥ BD and PS ∥ BD. Therefore QR ∥ PS.
4. Both pairs of opposite sides of PQRS are parallel. By the definition of a parallelogram, PQRS is a parallelogram.

Q6. Three parallel lines meet two transversals at A, B, C and D, E, F respectively, in corresponding order, with all six points distinct and AB = BC. Join AF, meeting the middle parallel BE at X. Prove DE = EF. [5 marks]

1. The given parallel relationships are AD ∥ BE ∥ CF. Since A, B and C lie in order on one transversal and AB = BC, B is the midpoint of AC.
2. In triangle ACF, BX lies on BE and is parallel to CF. The converse of the mid-point theorem therefore gives AX = XF, making X the midpoint of AF.
3. Now consider triangle FAD. The point X is the midpoint of its side FA, as just proved, and E lies on its side FD.
4. XE lies on the middle parallel BE. Since BE ∥ AD, the line through X and E is parallel to side AD of triangle FAD.
5. Apply the converse in triangle FAD: E is the midpoint of FD. Therefore DE = EF, proving that the second transversal has equal consecutive intercepts.

Q7. In triangle ABC, P, Q and R are the midpoints of AB, AC and BC respectively. Prove that triangles PQR and QPA are congruent by matching their three sides. [3 marks]

1. By the mid-point theorem, PQ = BC/2, QR = AB/2 and PR = AC/2. These give the three side lengths of triangle PQR.
2. The midpoint definitions give AP = AB/2 and AQ = AC/2. The side QP in triangle QPA is the same segment as PQ.
3. Thus PQ = QP, QR = PA and PR = QA when comparing the sides of the two triangles. Their three corresponding side lengths agree, so triangles PQR and QPA are congruent by SSS.

Key takeaways

• The mid-point theorem needs two midpoints and gives both parallelism and a joining segment half the third side's length.
• The converse needs one midpoint and a parallel line, and establishes the midpoint of another side.
• An auxiliary parallel line creates congruent triangles and a parallelogram in the proof of the mid-point theorem.
• State the triangle used at each stage when transferring midpoint information through a larger diagram.
• Joining the three side midpoints divides a triangle into four smaller triangles with equal corresponding side lengths.
• Joining successive side midpoints of a convex quadrilateral produces a parallelogram whose sides are parallel to the original diagonals.
• The equal intercept theorem transfers equality of consecutive intercepts from one transversal to another through the same parallel lines.
• Equal intercepts on different transversals need not equal each other; compare consecutive segments within each transversal.

Test yourself

What must be known before applying the direct mid-point theorem?

The endpoints of the joining segment must be the midpoints of two sides of the same triangle.

In triangle ABC, P and Q are the midpoints of AB and AC. Which side is parallel to PQ, and how are their lengths related?

PQ is parallel to BC, and its length is half that of BC: PQ = BC/2.

In triangle ABC, P is the midpoint of AB and PQ ∥ BC, with Q on AC. What does the converse establish?

It establishes that Q is the midpoint of AC, so AQ = QC.

Which fact allows the shorter proof of the converse to identify two constructed lines?

Through a point outside a given line, there is a unique line parallel to that line.

Which congruence test compares the four triangles formed by joining a triangle's side midpoints?

SSS applies because each small triangle has side lengths equal to halves of the original three sides.

What auxiliary segments help prove that the midpoint quadrilateral is a parallelogram?

The original quadrilateral's diagonals create triangles in which the mid-point theorem establishes the required parallel sides.

Three parallel lines meet transversals at A, B, C and D, E, F in corresponding order. If AB = BC, what follows?

The equal intercept theorem gives DE = EF, making E the midpoint of DF.

In that equal-intercept arrangement, must AB equal DE?

No. The theorem equates consecutive intercepts within each transversal, not the lengths on different transversals.