Exercise 8.4 — Mid Point Theorem

The mid point theorem of a triangle.

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Class 9 Mathematics · Chapter 8

Quadrilaterals — Exercise 8.4

Complete step-by-step solutions using the Midpoint Theorem — from finding BC when DE is given, to proving that midpoints of any quadrilateral form a parallelogram. Aligned with CBSE, Telangana, and Andhra Pradesh Class 9 syllabuses.

CBSE Class 9 Telangana Board Andhra Pradesh Board Chapter 8 · Ex 8.4

The Midpoint Theorem — Two Powerful Results

Exercise 8.4 is entirely driven by the Midpoint Theorem of a Triangle and its converse. Every question — from simple calculation to multi-step proof — is an application of one or both of these results. Knowing them cold is the key to solving this exercise quickly in board exams.

Midpoint Theorem

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.

In △ABC,
D = midpoint of AB
E = midpoint of BC
∴ DE ∥ AC and DE = ½AC

Converse

A line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side.

In △ABC,
D = midpoint of AB
DE ∥ BC
∴ AE = EC (E bisects AC)

Left: DE joins midpoints → DE ∥ BC, DE = ½BC. | Right: D is midpoint, DE ∥ BC → E bisects AC.

📐 Why two separate theorems? The first gives you a relationship between the midpoint segment and the third side. The second (converse) lets you locate a midpoint when you know a line is parallel to a side and passes through one midpoint. Both are used in this exercise — sometimes in the same question.

Question 1 — Finding BC When DE = 2 cm (Double Application)

Question 1

In △ABC, AD = ¼AB and AE = ¼AC. If DE = 2 cm, find BC.

The trick here is that D and E are quarter-way points, not midpoints. So you cannot apply the Midpoint Theorem to DE and BC directly. Instead, introduce the actual midpoints M (of AB) and N (of AC) and apply the theorem twice — once in △AMN and once in △ABC.

D, E are quarter-points; M, N are midpoints — the Midpoint Theorem is applied twice.

Let M = midpoint of AB, N = midpoint of AC. AD = ¼AB → AD = ½AM (since AM = ½AB) AE = ¼AC → AE = ½AN (since AN = ½AC) So D, E are midpoints of AM and AN respectively. Step 1: In △AMN, D and E are midpoints of AM and AN. By Midpoint Theorem: DE = ½ MN ← midpoint theorem in △AMN DE = 2 cm → MN = 2 × 2 = 4 cm Step 2: In △ABC, M and N are midpoints of AB and AC. By Midpoint Theorem: MN = ½ BC 4 = ½ BC → BC = 2 × 4 = 8 cm

✅ Answer: BC = 8 cm

💡 The double-step trick: Whenever a point is at ¼ or ¾ of a side (not the midpoint), introduce the actual midpoint and apply the Midpoint Theorem twice — once in the inner triangle, once in the outer triangle. This pattern appears frequently in board exam variations.

Question 2 — Midpoints of a Quadrilateral Form a Parallelogram

Question 2

ABCD is a quadrilateral. E, F, G, H are midpoints of AB, BC, CD, DA respectively. Prove that EFGH is a parallelogram.

The strategy is to introduce the diagonal AC. This divides the quadrilateral into two triangles — △ABC and △ADC — and the Midpoint Theorem applies to each.

Diagonal AC splits ABCD into △ABC and △ADC; Midpoint Theorem applied to each.

Draw diagonal AC.

Construction

In △ABC: E = midpoint of AB, F = midpoint of BC
∴ EF ∥ AC and EF = ½AC ...(1)

Midpoint Theorem in △ABC

In △ADC: H = midpoint of AD, G = midpoint of CD
∴ HG ∥ AC and HG = ½AC ...(2)

Midpoint Theorem in △ADC

From (1) and (2): EF ∥ HG and EF = HG

Both equal to ½AC and both parallel to AC

∴ EFGH is a parallelogram ✓

One pair of opposite sides parallel and equal

Question 3 — Midpoints of a Rhombus Form a Rectangle

Question 3

Show that the figure formed by joining the midpoints of the sides of a rhombus successively is a rectangle.

A rhombus has all sides equal and its diagonals are perpendicular to each other. This perpendicularity is the crucial extra condition that upgrades the inner figure from a mere parallelogram (proved in Q2) to a rectangle.

Rhombus ABCD with perpendicular diagonals at O; inner figure EFGH is a rectangle.

E, F, G, H are midpoints of AB, BC, CD, DA of rhombus ABCD.

Given

By Midpoint Theorem in △ABC: EF ∥ AC, EF = ½AC ...(1)
By Midpoint Theorem in △ADC: HG ∥ AC, HG = ½AC ...(2)

Midpoint Theorem in each triangle

EF ∥ HG and EF = HG → EFGH is a parallelogram

From (1) and (2) — one pair of opposite sides parallel and equal

Diagonals AC ⊥ BD of a rhombus → ∠AOB = 90°

Property of rhombus: diagonals are perpendicular

EF ∥ AC → ∠E = ∠AOB = 90°

Corresponding angles; EF is parallel to AC

Parallelogram EFGH has one angle = 90° → EFGH is a rectangle ✓

A parallelogram with one right angle is a rectangle

🔑 The key upgrade: Any quadrilateral's midpoints form a parallelogram (Q2). The rhombus adds perpendicular diagonals, which makes one angle of the inner parallelogram exactly 90° — turning it into a rectangle.

Question 4 — AF and EC Trisect Diagonal BD

Question 4

In parallelogram ABCD, E and F are midpoints of AB and DC. Show that AF and EC trisect diagonal BD.

This is a three-stage proof: first show AECF is a parallelogram, then use ASA congruence to get BQ = DP, then use the Midpoint Theorem converse in △DQC to get DP = PQ. Combining all three gives BP = PQ = QD.

AF and EC intersect diagonal BD at Q and P respectively; BP = PQ = QD (trisection).

Stage 1 — AECF is a parallelogram

AE = ½AB, CF = ½DC

E, F are midpoints of AB, DC

AB = DC → AE = CF and AE ∥ CF

Opposite sides of parallelogram ABCD

∴ AECF is a parallelogram

One pair of opposite sides parallel and equal

Stage 2 — △EBQ ≅ △FDP (so BQ = DP)

EB = FD (= ½AB = ½CD)

E, F are midpoints; AB = CD

∠EBQ = ∠FDP

Alternate interior angles (EB ∥ FD)

∠QEB = ∠PFD

∠QEB = ∠QCF = ∠PFD (corresponding / alternate angles)

△EBQ ≅ △FDP

ASA congruence rule

BQ = DP ...(1)

CPCT

Stage 3 — DP = PQ (Midpoint Theorem converse)

In △DQC: FP ∥ CQ and F = midpoint of DC

AECF is a parallelogram → FP ∥ CQ

By converse of Midpoint Theorem: P bisects DQ → DP = PQ ...(2)

Line through midpoint of DC parallel to CQ bisects DQ

From (1) and (2): DP = PQ = BQ → BD is trisected at P and Q ✓

Proved

Question 5 — Midpoints of Opposite Sides of a Quadrilateral Bisect Each Other

Question 5

Show that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.

E, F, G, H are midpoints of AB, BC, CD, DA. The "line segments joining opposite midpoints" means EG and FH. The goal is to show EG and FH bisect each other.

By Midpoint Theorem in △ABC: EF ∥ AC, EF = ½AC ...(1)

E = mid of AB, F = mid of BC

By Midpoint Theorem in △ADC: HG ∥ AC, HG = ½AC ...(2)

H = mid of DA, G = mid of CD

EF ∥ HG and EF = HG → EFGH is a parallelogram

One pair of opposite sides parallel and equal

Diagonals of parallelogram EFGH bisect each other.

Property of parallelogram: diagonals bisect each other

The diagonals of EFGH are EG and FH. ∴ EG and FH bisect each other ✓

Proved — the midpoints of opposite sides bisect each other

📌 Connection to Q2: Q5 is an extension of Q2. In Q2 you proved EFGH is a parallelogram. In Q5 you use that result and then apply the diagonal-bisection property of parallelograms to conclude EG and FH bisect each other.

Question 6 — Right Triangle: Line Through Midpoint of Hypotenuse

Question 6

ABC is a right-angled triangle (∠C = 90°). M is the midpoint of hypotenuse AB. MD ∥ BC intersects AC at D. Show that: (i) D is the midpoint of AC, (ii) MD ⊥ AC, (iii) CM = MA = ½AB.

M = midpoint of AB (hypotenuse); MD ∥ BC meets AC at D; CM = MA = ½AB is proved.

Part (i) — D is the midpoint of AC

M is midpoint of AB, MD ∥ BC

Given

By converse of Midpoint Theorem in △ABC:
MD bisects AC → AD = DC

Line through midpoint of AB, parallel to BC, bisects AC

∴ D is the midpoint of AC ✓

Proved (i)

Part (ii) — MD ⊥ AC

MD ∥ BC and ∠ACB = 90°

Given

∠ADM = ∠ACB = 90°

Corresponding angles (MD ∥ BC, AC is transversal)

∴ MD ⊥ AC ✓

Proved (ii)

Part (iii) — CM = MA = ½AB

In △ADM and △CDM:
AD = CD (D = midpoint of AC)
∠ADM = ∠CDM = 90°
DM = DM (common side)

Part (i) and (ii); common side

△ADM ≅ △CDM

SAS congruence rule

AM = CM ...(1)

CPCT

AM = ½AB ...(2)

M is midpoint of AB

From (1) and (2): CM = AM = ½AB ✓

Proved (iii)

✅ Geometric meaning of Part (iii): In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices. Here CM = AM = BM = ½AB — the midpoint of the hypotenuse is the circumcentre of a right triangle. This is a beautiful result that connects geometry and circle theorems.

Quick Summary — All 6 Questions

Q	Topic	Theorem / Rule Used	Result
1	D, E at ¼ of sides; DE = 2 cm	Midpoint Theorem applied twice	BC = 8 cm
2	Midpoints of a quadrilateral	Midpoint Theorem in △ABC and △ADC	EFGH is a parallelogram
3	Midpoints of a rhombus	Midpoint Theorem + perpendicular diagonals	EFGH is a rectangle
4	Midpoints of AB, DC in parallelogram	ASA congruence + Midpoint Theorem converse	AF, EC trisect BD
5	Midpoints of opposite sides of quadrilateral	Q2 result + diagonal bisection property	EG and FH bisect each other
6	Right triangle; M = midpoint of hypotenuse; MD ∥ BC	Midpoint Theorem converse + SAS congruence + CPCT	D = midpoint AC; MD ⊥ AC; CM = MA = ½AB

Common Mistakes to Avoid

Applying the Midpoint Theorem when points are not midpoints (Q1): DE joins quarter-points, not midpoints. You cannot write DE = ½BC directly. Always verify the points are midpoints before applying the theorem.
Forgetting to draw the diagonal in Q2 and Q3: The diagonal is the construction that allows the Midpoint Theorem to be applied to each triangle separately. Without stating "draw diagonal AC", the proof is incomplete.
Not distinguishing theorem from converse (Q4, Q6): The Midpoint Theorem says "midpoints → parallel and half". The converse says "one midpoint + parallel → bisects the third side". Using the wrong direction loses marks.
Missing the CPCT line in Q6(iii): After proving △ADM ≅ △CDM by SAS, you must explicitly write AM = CM by CPCT before combining with AM = ½AB.
Not explaining why EFGH is a rectangle (not just a parallelogram) in Q3: Many students prove EFGH is a parallelogram and stop. You must also prove ∠E = 90° using the perpendicularity of rhombus diagonals.

⛔ Board Exam Alert (CBSE / Telangana / AP): Q2 and Q3 are structurally very similar. In Q2 the answer is a parallelogram; in Q3 it becomes a rectangle because of the rhombus's perpendicular diagonals. If asked "what additional fact makes it a rectangle?", the answer is: diagonals of a rhombus are perpendicular.

What This Exercise Prepares You For

The Midpoint Theorem introduced in Exercise 8.4 is one of the most versatile tools in all of Class 9 and Class 10 geometry. It reappears in coordinate geometry when you derive the Midpoint Formula — the algebraic version of the same geometric idea. In Chapter 7 (Triangles), the same pattern of joining midpoints and applying parallel-line reasoning is used in the proof of the Basic Proportionality Theorem.

Q6(iii) — that the midpoint of the hypotenuse is equidistant from all three vertices — is the geometric root of the circumcircle theorem for right triangles, which surfaces explicitly in Class 10 circle theorems. Recognising these connections will help you link chapters rather than treating them in isolation.

📌 Exam Strategy (Telangana & AP): Q2–Q5 all involve a similar opening move: draw diagonal AC, apply the Midpoint Theorem in each triangle, conclude EF ∥ HG and EF = HG. Practise writing this three-line opening quickly — it is the shared entry point for all these proofs and knowing it cold saves time in timed exams.