Mid Point Theorem Class 9 ICSE — Maths Chapter 12
Mid Point Theorem Class 9 ICSE is a geometry chapter in the Class 9 Mathematics book. Chapter 12, Mid Point and Its Converse Including Intercept Theorem, teaches Theorem 6, its converse, and the equal intercept theorem. You will learn how each proof is built, then apply the results to triangles, trapeziums, and parallelograms. The chapter explains why each statement holds, so you can reproduce the proofs in your own words.
Download the Mid Point and Its Converse Including Intercept Theorem Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 11 pages, 2 sections, 31 questions.
What is inside Chapter 12
11 pages · 2 sections · 10 worked examples · 31 exercise questions
| Section | Title |
|---|---|
| 12.1 | MID-POINT THEOREM (Proof and simple applications and its converse) |
| 12.2 | EQUAL INTERCEPT THEOREM (Proof and simple application) |
- EXERCISE 12(A): 17 questions
- EXERCISE 12(B): 14 questions
12.1 MID-POINT THEOREM (Proof and simple applications and its converse)
Section 12.1 opens with Theorem 6. The mid point theorem states that the line joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. If D and E are the mid-points of AB and AC in triangle ABC, then DE \parallel BC and DE = \frac{1}{2}BC.
The proof uses one construction. Draw CF parallel to BA, meeting DE produced at F. In triangles ADE and CFE, AE = EC because E is the mid-point. Angles \angle AED and \angle CEF are vertically opposite; \angle EAD and \angle ECF are alternate. The triangles are congruent by ASA, so AD = CF. Since AD = BD, we get CF = BD.
Now CF and BD are equal and parallel, so BCFD is a parallelogram. Hence DE \parallel BC. Also DE = EF = \frac{1}{2}DF = \frac{1}{2}BC, so the two claims hold together.
The converse follows next. A straight line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side. The same construction proves it: with D the mid-point of AB and DE \parallel BC, triangles ADE and CFE are congruent by ASA, so AE = EC.
Worked example with original numbers: in triangle PQR, M and N are the mid-points of PQ and PR. If QR = 9.6 cm, then MN = 4.8 cm and MN \parallel QR. For the converse, a line through M parallel to QR meets PR at N. With PR = 12 cm, PN = NR = 6 cm.
The section then proves four standard results. The mid-points of the adjacent sides of any quadrilateral form a parallelogram, since each joining segment is parallel to the same diagonal and half its length. For a trapezium with AB \parallel DC, the mid-points of the non-parallel sides satisfy 2EF = AB + DC, proved by splitting along a diagonal.
The segment joining the mid-points of the diagonals of a trapezium satisfies EF = \frac{1}{2}(AB - DC). In a right triangle, the median to the hypotenuse equals half the hypotenuse, BD = \frac{1}{2}AC. These quadrilateral and trapezium results connect to the chapter on rectilinear figures.
12.2 EQUAL INTERCEPT THEOREM (Proof and simple application)
Section 12.2 states Theorem 8, the equal intercept theorem. If a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them also makes equal intercepts. With l \parallel m \parallel n and PQ = QR on the first transversal, the intercepts LM and MN on any other transversal are equal.
The proof has the same shape as the previous one. Draw PS and QT parallel to CD. In triangles PQS and QRT, PQ = QR is given. \angle PQS = \angle QRT and \angle QPS = \angle RQT are corresponding angles, because PS \parallel CD \parallel QT. The triangles are congruent by ASA, so PS = QT.
Now PSML and QTNM are parallelograms, giving PS = LM and QT = MN. Hence LM = MN. The theorem transfers equal spacing from one transversal to every other transversal.
The chapter applies the theorem three times. It re-proves the converse of the mid point theorem by drawing FG through A parallel to BC, so FG \parallel DE \parallel BC. The transversal AB makes equal intercepts AD = DB, so AC makes AE = CE.
In a parallelogram where E and F are the mid-points of AB and CD, any line crossing AD, EF, and BC gives GP = PH. When PB, AD, and QC are perpendicular to the same line, equal intercepts on PQ give BD = CD. SAS congruence of triangles ABD and ACD then gives AB = AC.
Worked example: three parallel lines cut by one transversal give equal intercepts of 2.5 cm each. Any second transversal therefore also gives intercepts of 2.5 cm each. Check the condition first: the intercepts on the first transversal must be equal before the theorem applies.
The table separates the three results by what is given and what you prove.
| Result | Given facts | What you prove |
|---|---|---|
| Mid-point theorem | D and E are mid-points of two sides of a triangle | DE \parallel BC and DE = \frac{1}{2}BC |
| Converse of mid-point theorem | D is the mid-point of AB, DE \parallel BC | E is the mid-point of AC, so AE = EC |
| Equal intercept theorem | l \parallel m \parallel n, PQ = QR on one transversal | LM = MN on any other transversal |
What to practise from Mid Point Theorem Class 9 ICSE
This chapter has two exercise blocks. Exercise 12(A) has 17 questions on Theorem 6 and its converse. Exercise 12(B) has 14 questions on the equal intercept theorem and mixed applications. The book’s main page lists all the Class 9 Maths chapters.
Exercise 12(A) covers direct length problems, such as finding AN and MN when BC and AC are given. It also practises the rectangle mid-points result and the trapezium result PR = \frac{1}{2}(AB + CD). Several questions combine congruence with the mid point theorem.
Exercise 12(B) practises reading intercepts from a figure and proving mid-points. It also uses parallel lines with equal intercepts to prove segment relations such as DS = 3RS or FB = \frac{1}{4}AB.
Common mistakes to avoid:
- The mid point theorem needs both endpoints to be mid-points of the triangle’s sides.
- The converse needs the parallel condition as well as the mid-point.
- The formula 2EF = AB + DC applies to the mid-points of the non-parallel sides.
- The formula EF = \frac{1}{2}(AB - DC) applies to the mid-points of the diagonals.
- Do not use the intercept theorem until equal intercepts on the first transversal are confirmed.
Exam pattern note: direct applications usually appear as short length or parallel questions. The trapezium and parallelogram results appear as proof questions. These map to the results most likely to appear in ICSE board tests. You can verify the syllabus on the official CISCE website.
Mid Point Theorem Class 9 ICSE frequently asked questions
What is the mid point theorem class 9 ICSE statement?
Theorem 6 states that the segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. In triangle ABC, with D and E the mid-points of AB and AC, DE \parallel BC and DE = \frac{1}{2}BC.
What is the converse of the mid point theorem?
The converse states that a line through the mid-point of one side of a triangle, parallel to another side, bisects the third side. So if D is the mid-point of AB and DE \parallel BC, then AE = EC.
What is the equal intercept theorem and when do I use it in Class 9 ICSE Maths?
The equal intercept theorem states that if a transversal makes equal intercepts on three or more parallel lines, any other transversal cutting those lines also makes equal intercepts. Use it when three parallel lines are given and you need to prove equal segments on a second transversal.
Why is the segment joining the mid points of the non parallel sides of a trapezium equal to half the sum of the parallel sides?
Draw diagonal BD, which splits the trapezium into two triangles. The mid point theorem gives half of AB in one triangle and half of DC in the other. Adding the halves gives 2EF = AB + DC.
How do I prove that joining the mid points of a quadrilateral makes a parallelogram?
Join one diagonal of the quadrilateral, say BD. In triangle ABD, the mid-points give PS \parallel BD and PS = \frac{1}{2}BD. In triangle BCD, they give QR \parallel BD and QR = \frac{1}{2}BD. One pair of opposite sides equal and parallel makes PQRS a parallelogram.
Reference: ICSE Class 9 Mathematics textbook material, chapter 12: Mid Point and Its Converse Including Intercept Theorem — CISCE-aligned.