Converse of Mid-Point Theorem
The Converse of the Mid-Point Theorem is a key result in the study of quadrilaterals and triangles in Class 9 NCERT Mathematics. While the Mid-Point Theorem connects midpoints of two sides to the third side, its converse works in the reverse direction.
The converse states: if a line is drawn through the midpoint of one side of a triangle parallel to another side, it will bisect the third side.
This theorem is used extensively in proving properties of parallelograms, trapeziums, and in construction problems. It is also known as the Intercept Theorem in some textbooks.
What is Converse of Mid-Point Theorem?
Mid-Point Theorem (for reference): The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half its length.
Converse of the Mid-Point Theorem:
If a line is drawn through the midpoint of one side of a triangle, parallel to another side, then it bisects the third side.
Formal statement:
- In ΔABC, let D be the midpoint of AB.
- If a line through D is drawn parallel to BC, meeting AC at E, then E is the midpoint of AC.
- That is, AE = EC.
Important:
- The line must pass through the midpoint of one side.
- The line must be parallel to another side.
- Both conditions must be met for the theorem to apply.
Converse of Mid-Point Theorem Formula
Key Results:
1. The Converse Statement:
In ΔABC, if D is the midpoint of AB and DE ∥ BC, then E is the midpoint of AC.
2. Length relationship:
DE = ½ × BC
3. Equal intercept theorem (extension):
- If a transversal makes equal intercepts on three or more parallel lines, then any other transversal cutting these lines will also make equal intercepts.
4. Application to trapezium:
- In a trapezium ABCD where AB ∥ DC, the line segment joining the midpoints of the non-parallel sides (AD and BC) is parallel to both parallel sides.
- Its length = ½(AB + DC).
Derivation and Proof
Proof of the Converse of the Mid-Point Theorem:
Given: In ΔABC, D is the midpoint of AB, and DE ∥ BC where E is on AC.
To prove: E is the midpoint of AC (AE = EC).
Construction: Through C, draw CF ∥ BA to meet DE produced at F.
Proof:
- DE ∥ BC (given) and CF ∥ BD (by construction).
- Therefore DBCF is a parallelogram (both pairs of opposite sides are parallel).
- In a parallelogram, opposite sides are equal: BD = CF. ... (i)
- D is the midpoint of AB (given), so AD = BD. ... (ii)
- From (i) and (ii): AD = CF. ... (iii)
Now consider ΔADE and ΔCFE:
- ∠AED = ∠CEF (vertically opposite angles)
- ∠ADE = ∠CFE (alternate interior angles, since AD ∥ CF and DF is the transversal)
- AD = CF (from (iii) above)
By AAS congruence: ΔADE ≅ ΔCFE
Therefore: AE = CE (CPCT)
Conclusion: E is the midpoint of AC. Hence proved.
Types and Properties
Applications and Extensions:
1. Finding midpoints in triangles:
- If we know the midpoint of one side and can draw a parallel to another side, the converse guarantees the third side is also bisected.
- Useful in constructions where direct measurement is difficult.
2. Equal Intercept Theorem:
- If three parallel lines l, m, n make equal intercepts on a transversal p, they make equal intercepts on every transversal q.
- This generalises the converse of the mid-point theorem.
3. Properties of parallelogram diagonals:
- The diagonals of a parallelogram bisect each other. This can be established using the mid-point theorem and its converse.
4. Trapezium midsegment:
- The segment joining the midpoints of the non-parallel sides of a trapezium is parallel to the parallel sides and equals half their sum.
- This is proved using the converse of the mid-point theorem.
5. Quadrilateral formed by joining midpoints:
- The quadrilateral formed by joining the midpoints of the sides of any quadrilateral is always a parallelogram (Varignon's Theorem).
- This is proved by applying the mid-point theorem to the two triangles formed by a diagonal.
Solved Examples
Example 1: Example 1: Basic application of the converse
Problem: In ΔPQR, S is the midpoint of PQ and ST ∥ QR. If PR = 12 cm, find PT.
Solution:
Given:
- S is the midpoint of PQ.
- ST ∥ QR, where T lies on PR.
- PR = 12 cm.
By the Converse of the Mid-Point Theorem:
- Since S is the midpoint of PQ and ST ∥ QR, T is the midpoint of PR.
- PT = PR / 2 = 12 / 2 = 6 cm
Answer: PT = 6 cm.
Example 2: Example 2: Finding the length of the segment
Problem: In ΔABC, D is the midpoint of AB and DE ∥ BC. If BC = 14 cm, find DE.
Solution:
Given:
- D is the midpoint of AB.
- DE ∥ BC.
- BC = 14 cm.
By the Mid-Point Theorem:
- Since D is the midpoint of AB and DE ∥ BC (which means E is the midpoint of AC by the converse), DE = ½ × BC.
- DE = ½ × 14 = 7 cm
Answer: DE = 7 cm.
Example 3: Example 3: Prove that a point is a midpoint
Problem: In ΔXYZ, M is the midpoint of XY. A line through M parallel to YZ meets XZ at N. Prove that N is the midpoint of XZ.
Solution:
Given:
- M is the midpoint of XY (XM = MY).
- MN ∥ YZ, where N is on XZ.
Proof:
- In ΔXYZ, M is the midpoint of XY (given).
- MN ∥ YZ (given).
- By the Converse of the Mid-Point Theorem, the line through the midpoint of one side, parallel to another side, bisects the third side.
- Therefore, N is the midpoint of XZ.
- That is, XN = NZ.
Hence proved.
Example 4: Example 4: Trapezium midsegment
Problem: In trapezium ABCD, AB ∥ DC, AB = 16 cm and DC = 10 cm. E and F are the midpoints of AD and BC respectively. Find EF.
Solution:
Given:
- AB ∥ DC
- AB = 16 cm, DC = 10 cm
- E = midpoint of AD, F = midpoint of BC
Using the midsegment theorem for a trapezium:
- EF ∥ AB ∥ DC
- EF = ½(AB + DC) = ½(16 + 10) = ½ × 26 = 13 cm
Answer: EF = 13 cm.
Example 5: Example 5: Using the converse in a quadrilateral
Problem: ABCD is a quadrilateral. P, Q, R, S are the midpoints of AB, BC, CD, DA respectively. Prove that PQRS is a parallelogram.
Solution:
Construction: Join diagonal AC.
In ΔABC:
- P is the midpoint of AB, Q is the midpoint of BC.
- By the Mid-Point Theorem: PQ ∥ AC and PQ = ½ AC. ... (i)
In ΔACD:
- S is the midpoint of AD, R is the midpoint of CD.
- By the Mid-Point Theorem: SR ∥ AC and SR = ½ AC. ... (ii)
From (i) and (ii):
- PQ ∥ SR and PQ = SR
- One pair of opposite sides is both parallel and equal.
- Therefore, PQRS is a parallelogram.
Hence proved.
Example 6: Example 6: Finding coordinates using the converse
Problem: In ΔABC, A = (0, 0), B = (6, 0), C = (4, 8). D is the midpoint of AB. A line through D parallel to BC meets AC at E. Find E.
Solution:
Find D (midpoint of AB):
- D = ((0+6)/2, (0+0)/2) = (3, 0)
By the Converse of the Mid-Point Theorem:
- E is the midpoint of AC.
- E = ((0+4)/2, (0+8)/2) = (2, 4)
Verification: Slope of BC = (8−0)/(4−6) = 8/(−2) = −4. Slope of DE = (4−0)/(2−3) = 4/(−1) = −4. Since slopes are equal, DE ∥ BC. ✓
Answer: E = (2, 4).
Example 7: Example 7: Equal intercept theorem application
Problem: Three parallel lines l, m, n cut a transversal at A, B, C such that AB = BC = 5 cm. Another transversal cuts them at P, Q, R. If PQ = 7 cm, find QR.
Solution:
Given:
- l ∥ m ∥ n
- AB = BC = 5 cm (equal intercepts on one transversal)
By the Equal Intercept Theorem:
- If three parallel lines make equal intercepts on one transversal, they make equal intercepts on any other transversal.
- Therefore, PQ = QR.
- QR = 7 cm.
Answer: QR = 7 cm.
Example 8: Example 8: Finding sides of a triangle
Problem: In ΔABC, AB = 10 cm, BC = 14 cm, AC = 12 cm. D, E, F are the midpoints of AB, BC, AC respectively. Find the perimeter of ΔDEF.
Solution:
By the Mid-Point Theorem:
- DE ∥ AC and DE = ½ × AC = ½ × 12 = 6 cm
- EF ∥ AB and EF = ½ × AB = ½ × 10 = 5 cm
- DF ∥ BC and DF = ½ × BC = ½ × 14 = 7 cm
Perimeter of ΔDEF:
- = DE + EF + DF = 6 + 5 + 7 = 18 cm
Answer: The perimeter of ΔDEF is 18 cm.
Example 9: Example 9: Converse in a right triangle
Problem: In a right triangle PQR with ∠Q = 90°, PQ = 8 cm and QR = 6 cm. M is the midpoint of PR. A line through M parallel to QR meets PQ at N. Find NQ.
Solution:
First find PR:
- PR = √(PQ² + QR²) = √(64 + 36) = √100 = 10 cm
By the Converse of the Mid-Point Theorem:
- M is the midpoint of PR and MN ∥ QR.
- Therefore, N is the midpoint of PQ.
- NQ = PQ / 2 = 8 / 2 = 4 cm
Answer: NQ = 4 cm.
Example 10: Example 10: Multi-step problem
Problem: In ΔABC, D and E are the midpoints of AB and AC. DE is extended to F such that DE = EF. Prove that DBCF is a parallelogram.
Solution:
Given: D is midpoint of AB, E is midpoint of AC, DE = EF.
Proof:
- By the Mid-Point Theorem, DE ∥ BC and DE = ½ BC. ... (i)
- DE = EF (given), so DF = 2 × DE = BC. ... (ii)
- Since DE ∥ BC and F is on DE produced, DF ∥ BC.
Now consider the quadrilateral DBCF:
- From (i) and extension: DF ∥ BC ... (iii)
- From (ii): DF = BC ... (iv)
Since one pair of opposite sides (DF and BC) is both parallel and equal, DBCF is a parallelogram.
Hence proved.
Real-World Applications
Applications of the Converse of the Mid-Point Theorem:
- Proving properties of quadrilaterals: Used to establish that the figure formed by joining midpoints of any quadrilateral is always a parallelogram.
- Trapezium properties: The midsegment of a trapezium (line joining midpoints of non-parallel sides) is proved using this theorem.
- Construction: Bisecting a line segment by drawing a parallel through a known midpoint, without using a compass.
- Coordinate geometry: Finding unknown vertices or midpoints of triangles when some coordinates are given.
- Engineering and design: Structural analysis of trusses and frameworks often uses midpoint properties for load distribution.
- Map reading: Determining midpoints and proportional distances on maps using parallel grid lines and the intercept theorem.
Key Points to Remember
- The Converse of the Mid-Point Theorem states: a line through the midpoint of one side of a triangle, drawn parallel to another side, bisects the third side.
- Both conditions are needed: midpoint of one side AND parallel to another side.
- The line segment DE (joining midpoints) is half the length of the third side BC.
- The Equal Intercept Theorem generalises this: equal intercepts on one transversal imply equal intercepts on every transversal.
- The quadrilateral formed by joining midpoints of any quadrilateral is a parallelogram (Varignon’s Theorem).
- The midsegment of a trapezium with parallel sides a and b has length (a + b)/2.
- The converse is used to prove that a point is a midpoint, given the parallel condition.
- The proof uses AAS congruence and properties of parallel lines.
- This theorem is part of the Quadrilaterals chapter in CBSE Class 9.
- Combined with the Mid-Point Theorem, it provides a complete toolkit for midpoint-based proofs.
Practice Problems
- In ΔABC, D is the midpoint of AB. DE ∥ BC meets AC at E. If AC = 18 cm, find AE.
- In ΔPQR, M is the midpoint of PQ and MN ∥ QR. If QR = 20 cm, find MN.
- ABCD is a trapezium with AB ∥ DC. AB = 20 cm and DC = 8 cm. Find the length of the line joining the midpoints of AD and BC.
- In ΔXYZ, A is the midpoint of XY. A line through A parallel to YZ meets XZ at B. If XZ = 15 cm, find XB and BZ.
- Prove that the line drawn through the midpoint of one side of a triangle, parallel to the base, bisects the other side.
- ABCD is a quadrilateral. P, Q, R, S are midpoints of AB, BC, CD, DA. If AC = 10 cm and BD = 12 cm, find the perimeter of PQRS.
- In ΔABC, D is the midpoint of BC. DE ∥ AB meets AC at E. Prove that E is the midpoint of AC.
- Three parallel lines make intercepts of 3 cm each on one transversal. They cut another transversal at P, Q, R. If PQ = 4.5 cm, find QR.
Frequently Asked Questions
Q1. What is the Converse of the Mid-Point Theorem?
In a triangle, if a line is drawn through the midpoint of one side parallel to another side, it bisects the third side. That is, the point where it meets the third side is also a midpoint.
Q2. How is the converse different from the Mid-Point Theorem?
The Mid-Point Theorem starts with midpoints of two sides and concludes the joining segment is parallel to the third side. The converse starts with a midpoint and a parallel condition and concludes the third side is bisected.
Q3. What is the Equal Intercept Theorem?
If three or more parallel lines make equal intercepts on one transversal, they make equal intercepts on any other transversal. It is a generalisation of the converse of the mid-point theorem.
Q4. Can the converse be used for quadrilaterals?
Yes. It is used to prove that joining midpoints of any quadrilateral gives a parallelogram (Varignon's Theorem), and to find the midsegment of a trapezium.
Q5. What are the conditions for the converse to apply?
Two conditions must hold: (1) the line passes through the midpoint of one side of the triangle, and (2) the line is parallel to another side. If both are met, the third side is bisected.
Q6. What is the length of the midsegment in a trapezium?
In a trapezium with parallel sides of lengths a and b, the midsegment (line joining midpoints of non-parallel sides) has length (a + b)/2.
Q7. How is this theorem proved?
By constructing a line through C parallel to AB, forming a parallelogram, and then using AAS congruence to show the point on AC is its midpoint.
Q8. Is this topic in CBSE Class 9?
Yes. The Converse of the Mid-Point Theorem is part of Chapter 8 (Quadrilaterals) in the CBSE Class 9 NCERT Mathematics textbook.
