Median Divides Triangle into Equal Areas
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Every triangle has exactly three medians.
One of the most important results in Class 9 Geometry states that a median of a triangle divides it into two triangles of equal area. This holds true for all types of triangles — scalene, isosceles, and equilateral.
This result follows directly from the fact that the two smaller triangles share the same base length (since the median bisects the opposite side) and have the same height (the perpendicular distance from the common vertex to the base line).
What is Median Divides Triangle into Equal Areas?
Definition: A median of a triangle is a line segment drawn from any vertex to the midpoint of the opposite side.
Theorem: A median of a triangle divides it into two triangles of equal area.
If AD is a median of triangle ABC, where D is the midpoint of BC, then:
Area(▵ABD) = Area(▵ACD) = ½ × Area(▵ABC)
Where:
- D is the midpoint of side BC, so BD = DC
- AD is the median from vertex A
- Both triangles ABD and ACD share the same vertex A
Median Divides Triangle into Equal Areas Formula
Key Formulas:
1. Area of each sub-triangle:
Area(▵ABD) = Area(▵ACD) = ½ × Area(▵ABC)
2. Area using base and height:
- Let h be the perpendicular from A to line BC.
- Area(▵ABD) = ½ × BD × h
- Area(▵ACD) = ½ × DC × h
- Since BD = DC, both areas are equal.
3. If all three medians are drawn:
- The three medians divide the triangle into 6 smaller triangles of equal area.
- Each small triangle has area = ⅙ × Area(▵ABC).
Derivation and Proof
Proof that a median divides a triangle into two triangles of equal area:
Given: In ▵ABC, AD is a median where D is the midpoint of BC.
To Prove: Area(▵ABD) = Area(▵ACD)
Construction: Draw AE ⊥ BC (or BC extended).
Proof:
- Since D is the midpoint of BC, we have BD = DC.
- The height from A to line BC is the same for both triangles — it is AE = h.
- Area(▵ABD) = ½ × BD × h
- Area(▵ACD) = ½ × DC × h
- Since BD = DC, we get Area(▵ABD) = Area(▵ACD).
Also:
- Area(▵ABD) + Area(▵ACD) = Area(▵ABC)
- Since both are equal: Area(▵ABD) = Area(▵ACD) = ½ × Area(▵ABC)
Hence proved.
Types and Properties
Related Results:
1. All Three Medians
- Every triangle has three medians: from A to midpoint of BC, from B to midpoint of AC, from C to midpoint of AB.
- Each median divides the triangle into two equal-area triangles.
2. Centroid and Six Sub-Triangles
- The three medians are concurrent at the centroid G.
- The three medians divide the triangle into six smaller triangles of equal area.
- Each small triangle has area = (1/6) × Area(▵ABC).
3. Centroid Division Property
- The centroid divides each median in the ratio 2 : 1 from vertex to midpoint.
- AG : GD = 2 : 1, where G is the centroid and D is the midpoint of BC.
4. Converse Application
- If a line segment from a vertex divides the opposite side into two equal parts, the areas of the two resulting triangles are equal.
- This is used in coordinate geometry to verify midpoints.
Solved Examples
Example 1: Example 1: Verify equal areas using base and height
Problem: In ▵ABC, the area is 48 sq cm. D is the midpoint of BC. Find the area of ▵ABD and ▵ACD.
Solution:
Given:
- Area(▵ABC) = 48 sq cm
- D is the midpoint of BC, so AD is a median.
Using the theorem:
- Area(▵ABD) = ½ × 48 = 24 sq cm
- Area(▵ACD) = ½ × 48 = 24 sq cm
Answer: Each triangle has area 24 sq cm.
Example 2: Example 2: Finding area with given side and height
Problem: In ▵PQR, PS is a median to QR. If QR = 12 cm and the height from P to QR is 8 cm, find the area of ▵PQS.
Solution:
Given:
- QR = 12 cm, so QS = SR = 6 cm (S is midpoint)
- Height from P to QR = 8 cm
Finding the area:
- Area(▵PQR) = ½ × 12 × 8 = 48 sq cm
- Area(▵PQS) = ½ × Area(▵PQR) = ½ × 48 = 24 sq cm
Verification:
- Area(▵PQS) = ½ × QS × h = ½ × 6 × 8 = 24 sq cm ✓
Answer: Area(▵PQS) = 24 sq cm.
Example 3: Example 3: Coordinate geometry application
Problem: The vertices of ▵ABC are A(0, 6), B(0, 0), C(8, 0). D is the midpoint of BC. Verify that AD divides the triangle into equal areas.
Solution:
Given:
- A(0, 6), B(0, 0), C(8, 0)
- D = midpoint of BC = ((0+8)/2, (0+0)/2) = (4, 0)
Area of ▵ABC:
- = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
- = ½ |0(0 − 0) + 0(0 − 6) + 8(6 − 0)| = ½ × 48 = 24 sq units
Area of ▵ABD:
- = ½ |0(0 − 0) + 0(0 − 6) + 4(6 − 0)| = ½ × 24 = 12 sq units
Area of ▵ACD:
- = ½ |0(0 − 0) + 8(0 − 6) + 4(6 − 0)| = ½ |−48 + 24| = ½ × 24 = 12 sq units
Answer: Area(▵ABD) = Area(▵ACD) = 12 sq units, which is ½ × 24. Verified.
Example 4: Example 4: Using the six equal triangles property
Problem: In ▵ABC, all three medians are drawn, meeting at centroid G. If the area of ▵ABC is 72 sq cm, find the area of each of the six smaller triangles.
Solution:
Given:
- Area(▵ABC) = 72 sq cm
- Three medians create six sub-triangles at the centroid.
Using the property:
- Each of the 6 triangles has equal area.
- Area of each = 72 / 6 = 12 sq cm
Answer: Each smaller triangle has area 12 sq cm.
Example 5: Example 5: Proving ratio of areas
Problem: In ▵ABC, D is the midpoint of BC and E is the midpoint of AD. Prove that Area(▵ABE) = ¼ × Area(▵ABC).
Solution:
Given:
- D is the midpoint of BC, E is the midpoint of AD.
Proof:
- AD is a median of ▵ABC. So Area(▵ABD) = ½ × Area(▵ABC).
- In ▵ABD, BE is a median (E is the midpoint of AD).
- So Area(▵ABE) = ½ × Area(▵ABD).
- = ½ × ½ × Area(▵ABC) = ¼ × Area(▵ABC).
Answer: Area(▵ABE) = ¼ × Area(▵ABC). Proved.
Example 6: Example 6: Numerical problem with obtuse triangle
Problem: An obtuse triangle has sides 13 cm, 14 cm, and 15 cm. AD is the median to the side of length 14 cm. Find the area of each sub-triangle.
Solution:
Given:
- a = 14 cm (base), b = 13 cm, c = 15 cm
Finding area using Heron's Formula:
- s = (13 + 14 + 15) / 2 = 21
- Area = √(21 × 8 × 7 × 6) = √7056 = 84 sq cm
Area of each sub-triangle:
- Area(▵ABD) = Area(▵ACD) = 84 / 2 = 42 sq cm
Answer: Each sub-triangle has area 42 sq cm.
Example 7: Example 7: Finding area when centroid is given
Problem: In ▵PQR, the centroid G divides median PS such that PG = 8 cm and GS = 4 cm. If QR = 10 cm and the height from P to QR is 12 cm, find the area of ▵QGR.
Solution:
Given:
- PG = 8 cm, GS = 4 cm (ratio 2:1, confirming G is centroid)
- QR = 10 cm, height from P = 12 cm
Finding areas:
- Area(▵PQR) = ½ × 10 × 12 = 60 sq cm
- The centroid divides the triangle into 6 equal sub-triangles.
- Area of each = 60 / 6 = 10 sq cm
- ▵QGR consists of 2 such sub-triangles (▵QGS and ▵GRS).
- Area(▵QGR) = 2 × 10 = 20 sq cm
Answer: Area(▵QGR) = 20 sq cm.
Example 8: Example 8: Word problem
Problem: A triangular field ABC has area 600 sq m. A farmer marks the midpoint D of side BC and fences the region ABD to grow wheat. What area is used for wheat?
Solution:
Given:
- Area(▵ABC) = 600 sq m
- D is the midpoint of BC, so AD is a median.
Using the theorem:
- The median AD divides ▵ABC into two triangles of equal area.
- Area(▵ABD) = ½ × 600 = 300 sq m
Answer: The farmer uses 300 sq m for wheat.
Real-World Applications
Applications:
- Land Division: Dividing a triangular plot into two equal areas by marking the midpoint of one side and drawing a line to the opposite vertex.
- Centre of Gravity: The centroid (intersection of medians) is the centre of mass of a uniform triangular plate. Each median line is a balance axis.
- Coordinate Geometry: Verifying midpoints and area relationships using the coordinate formula for area of a triangle.
- Proof Technique: The median-area property is used as a building block in many advanced geometry proofs, especially those involving equal areas and ratios.
- Structural Engineering: Trusses and triangular frames use centroid properties for load distribution calculations.
Key Points to Remember
- A median joins a vertex to the midpoint of the opposite side.
- Every triangle has three medians.
- A median divides the triangle into two triangles of equal area.
- The proof relies on both sub-triangles having the same base (BD = DC) and the same height.
- The three medians are concurrent at the centroid.
- The centroid divides each median in the ratio 2 : 1 from vertex to midpoint.
- All three medians divide the triangle into 6 triangles of equal area.
- Each of the 6 sub-triangles has area = (1/6) × area of the original triangle.
- This property holds for all types of triangles — scalene, isosceles, equilateral, acute, obtuse, and right-angled.
- The result is frequently used in coordinate geometry and proof-based problems in NCERT Class 9.
Practice Problems
- In triangle ABC, area is 96 sq cm. M is the midpoint of BC. Find the area of triangle ABM.
- The vertices of triangle PQR are P(2, 5), Q(4, 1), R(8, 3). S is the midpoint of QR. Verify that PS divides the triangle into two equal areas.
- In triangle DEF, all three medians meet at G. If the area of triangle DEF is 108 sq cm, find the area of triangle DGE.
- AD is a median of triangle ABC. E is the midpoint of AD. Show that area of triangle BEC = half the area of triangle ABC.
- A triangular garden has vertices at A(0, 0), B(10, 0), C(4, 8). D is the midpoint of AB. Find the area of triangle ACD.
- In triangle XYZ, the area is 150 sq cm. M and N are midpoints of XY and XZ respectively. Find the area of triangle XMN.
Frequently Asked Questions
Q1. What is a median of a triangle?
A median is a line segment joining a vertex of a triangle to the midpoint of the opposite side. Every triangle has three medians, and they all meet at a single point called the centroid.
Q2. How does a median divide a triangle into equal areas?
The median from vertex A to midpoint D of BC creates two triangles ABD and ACD. Both have the same base length (BD = DC) and the same height (perpendicular from A to BC). Since area = ½ × base × height, and base and height are equal, the areas are equal.
Q3. Do all three medians divide a triangle into equal parts?
Yes. The three medians divide a triangle into six smaller triangles. All six have equal area. Each small triangle has area = (1/6) × area of the original triangle.
Q4. What is the centroid of a triangle?
The centroid is the point where all three medians intersect. It divides each median in the ratio 2:1 from the vertex to the midpoint of the opposite side. It is also the centre of mass of a uniform triangular lamina.
Q5. Does this property hold for all types of triangles?
Yes. The median divides any triangle into two triangles of equal area, regardless of whether it is scalene, isosceles, equilateral, acute, obtuse, or right-angled.
Q6. Can this theorem be used in coordinate geometry?
Yes. If vertices are given as coordinates, find the midpoint using the midpoint formula. Then use the coordinate area formula to verify that the two sub-triangles have equal areas.
Q7. Is this theorem in the CBSE Class 9 syllabus?
Yes. This result is covered in Chapter 9 (Areas of Parallelograms and Triangles) of NCERT Class 9 Mathematics. It is frequently asked in exams.
Q8. What is the difference between a median and an altitude?
A median joins a vertex to the midpoint of the opposite side. An altitude is a perpendicular from a vertex to the opposite side (or its extension). A median always lies inside the triangle; an altitude may lie outside in an obtuse triangle.
