Mid-Point Theorem and Pythagoras Theorem
Mid-Point Theorem
Statement: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
Proof
Given: In triangle ABC, D is the mid-point of AB, and E is the mid-point of AC.
To prove: DE is parallel to BC and DE = 1/2 × BC
Construction: Extend DE to F such that DE = EF. Join CF.
Proof:
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In triangles ADE and CFE:
- AE = EC (E is mid-point of AC)
- DE = EF (by construction)
- Angle AED = Angle CEF (vertically opposite angles)
- Therefore, triangle ADE is congruent to triangle CFE (SAS)
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By CPCT: AD = CF and Angle ADE = Angle CFE
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Since Angle ADE = Angle CFE (alternate interior angles are equal), AD is parallel to CF.
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AD = DB (D is mid-point) and AD = CF (proved), so DB = CF.
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In quadrilateral DBCF, DB is parallel to CF and DB = CF. Therefore, DBCF is a parallelogram.
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DF is parallel to BC and DF = BC.
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Since DE = 1/2 × DF, we have DE = 1/2 × BC.
Hence proved: DE is parallel to BC and DE = 1/2 × BC.
Applications
<ICSEExample title="Mid-Point Theorem"> In triangle ABC, D and E are mid-points of AB and AC respectively. If BC = 10 cm, find DE. <Solution> By mid-point theorem: DE = 1/2 × BC DE = 1/2 × 10 = 5 cm Also, DE is parallel to BC. </Solution> </ICSEExample>Converse of Mid-Point Theorem
Statement: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.
Pythagoras Theorem
Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
If ABC is a right triangle with right angle at B: AC² = AB² + BC²
Proof by Area Method
Given: Right triangle ABC with right angle at B.
Construction: Draw squares on each side: square ABDE on AB, square BCFG on BC, and square ACHI on AC.
Proof: Consider a right triangle ABC. Construct a square of side (a + b) where a = BC, b = AB.
The area of the large square = (a + b)² = a² + 2ab + b²
This area equals the area of 4 right triangles (each area = ab/2) plus the area of the inner square (side c, area c²).
Therefore: a² + 2ab + b² = 4(ab/2) + c² a² + 2ab + b² = 2ab + c² a² + b² = c²
Hence, the square of the hypotenuse equals sum of squares of the other two sides.
<ICSEExample title="Pythagoras Theorem"> Find the hypotenuse of a right triangle with sides 6 cm and 8 cm. <Solution> Let hypotenuse = h h² = 6² + 8² = 36 + 64 = 100 h = 10 cm </Solution> </ICSEExample> <ICSEExample title="Application of Pythagoras"> A ladder 13 m long reaches a window 12 m above the ground. Find the distance of the foot of the ladder from the wall. <Solution> Let distance = d 13² = 12² + d² d² = 169 - 144 = 25 d = 5 m </Solution> </ICSEExample>Pythagorean Triplets
A Pythagorean triplet is a set of three positive integers a, b, c such that a² + b² = c².
Common Triplets: (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (12, 16, 20)
Generating Triplets: For m > n, (m² - n², 2mn, m² + n²) forms a Pythagorean triplet.
Common Mistakes With Fixes
| Mistake | Correction |
|---|---|
| Applying mid-point theorem to non-triangle figures | Mid-point theorem applies only to triangles |
| Confusing which side is half of which | DE is half of BC, not vice versa |
| Wrong identification of hypotenuse | Hypotenuse is the side opposite the right angle |
| Forgetting units in Pythagoras problems | Always include units in the final answer |
ICSE Exam Focus
| Topic | Marks (approx.) | Frequency |
|---|---|---|
| Mid-point theorem (proof and application) | 4-5 marks | Very common |
| Pythagoras theorem (direct application) | 3-4 marks | Very common |
| Word problems using Pythagoras | 4 marks | Common |
| Converse of mid-point theorem | 3 marks | Occasionally asked |
Self-Test
Q1: In triangle ABC, D and E are mid-points of AB and AC. If DE = 6 cm, find BC.
Q2: A right triangle has sides 9 cm and 12 cm. Find the hypotenuse.
Q3: Find the distance between two points A(3, 4) and B(6, 8) using Pythagoras.
Q4: The diagonals of a rhombus are 16 cm and 12 cm. Find its side length.
Q5: Prove the mid-point theorem.
