Triangles on Same Base and Between Same Parallels
In NCERT Class 9 Mathematics, the chapter on Areas of Parallelograms and Triangles establishes a fundamental result: triangles on the same base and between the same parallels have equal area.
This theorem connects the concept of area with the position of the vertex relative to the base. As long as the vertex moves along a line parallel to the base, the area of the triangle remains unchanged.
This result has wide applications in proving area relationships, solving construction problems, and understanding properties of quadrilaterals.
What is Triangles on Same Base and Between Same Parallels?
Theorem: Triangles on the same base (or equal bases) and between the same parallels are equal in area.
If ΔABC and ΔDBC are on the same base BC and between the same parallels BC and AD, then ar(ΔABC) = ar(ΔDBC).
Why this works:
- Area of a triangle = ½ × base × height.
- Both triangles share the same base BC.
- Since A and D lie on a line parallel to BC, the perpendicular height from A to BC equals the perpendicular height from D to BC.
- Same base and same height ⇒ same area.
Converse: If two triangles have the same base and equal areas, their vertices lie on a line parallel to the base (i.e., they lie between the same parallels).
Triangles on Same Base and Between Same Parallels Formula
Key Results:
1. Equal area theorem:
ar(ΔABC) = ar(ΔDBC), if BC is the common base and AD ∥ BC
2. Area formula:
ar(Δ) = ½ × base × height
3. Related result for parallelograms:
- A diagonal of a parallelogram divides it into two triangles of equal area.
- If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is half the area of the parallelogram.
4. Median divides triangle:
- A median of a triangle divides it into two triangles of equal area.
- This is a direct consequence of the same-base-same-parallels theorem.
Derivation and Proof
Proof that triangles on the same base and between the same parallels have equal area:
Given: ΔABC and ΔDBC lie on the same base BC, and vertices A and D lie on a line ℓ ∥ BC.
To prove: ar(ΔABC) = ar(ΔDBC).
Proof:
- Draw AE ⊥ BC and DF ⊥ BC, where E and F are the feet of the perpendiculars on BC (or BC produced).
- Since ℓ ∥ BC, the perpendicular distance between the two parallel lines is constant.
- Therefore, AE = DF = h (the distance between the parallels). ... (i)
- ar(ΔABC) = ½ × BC × AE = ½ × BC × h ... (ii)
- ar(ΔDBC) = ½ × BC × DF = ½ × BC × h ... (iii)
- From (ii) and (iii): ar(ΔABC) = ar(ΔDBC).
Hence proved.
Proof of the converse:
Given: ΔABC and ΔDBC on the same base BC such that ar(ΔABC) = ar(ΔDBC).
To prove: AD ∥ BC.
- ar(ΔABC) = ½ × BC × h₁ (where h₁ is the height from A to BC)
- ar(ΔDBC) = ½ × BC × h₂ (where h₂ is the height from D to BC)
- Since the areas are equal: ½ × BC × h₁ = ½ × BC × h₂
- Therefore h₁ = h₂.
- A and D are at the same perpendicular distance from BC, and on the same side.
- Therefore AD ∥ BC.
Types and Properties
Related Theorems and Extensions:
1. Triangle and parallelogram on the same base:
- If a triangle and a parallelogram share the same base and lie between the same parallels, the area of the triangle is half the area of the parallelogram.
- ar(Δ) = ½ × ar(parallelogram)
2. Parallelograms on the same base and between the same parallels:
- They are equal in area.
- Different shapes, same area — the key insight.
3. Median theorem:
- A median divides a triangle into two triangles of equal area.
- Proof: Both triangles have the same height (from the vertex to the base) and equal bases (the median bisects the base).
4. Triangles on equal bases and between the same parallels:
- Even if the bases are not the same segment but have the same length, and the triangles lie between the same parallels, the areas are equal.
5. Diagonal of a trapezium:
- In a trapezium ABCD (AB ∥ DC), the diagonal AC creates ΔABC and ΔACD.
- ΔABD and ΔABC share base AB and lie between the same parallels AB and DC.
- Therefore ar(ΔABD) = ar(ΔABC).
Solved Examples
Example 1: Example 1: Equal area triangles
Problem: ΔABC and ΔDBC are on the same base BC. A and D are on the same side of BC such that AD ∥ BC. If BC = 10 cm and the distance between BC and AD is 6 cm, find the area of each triangle.
Solution:
Given:
- BC = 10 cm (common base)
- Height (distance between parallels) = 6 cm
Area of each triangle:
- ar(ΔABC) = ½ × 10 × 6 = 30 sq cm
- ar(ΔDBC) = ½ × 10 × 6 = 30 sq cm
Answer: Each triangle has area = 30 sq cm.
Example 2: Example 2: Prove areas equal
Problem: In the figure, ABCD is a parallelogram. P is any point on AB. Prove that ar(ΔDPC) = ar(ΔAPD) + ar(ΔBPC).
Solution:
In parallelogram ABCD:
- AB ∥ DC and AB = DC.
Consider ΔDPC and parallelogram ABCD:
- ar(ABCD) = AB × h (where h is the height).
- ΔDPC has base DC = AB, and vertex P lies on AB (which is parallel to DC).
- Height of ΔDPC from P to DC = h (distance between AB and DC).
- ar(ΔDPC) = ½ × DC × h = ½ × ar(ABCD). ... (i)
Also:
- ar(ΔAPD) + ar(ΔDPC) + ar(ΔBPC) = ar(ABCD). ... (ii)
From (i) and (ii): ar(ΔAPD) + ar(ΔBPC) = ar(ABCD) − ar(ΔDPC) = ar(ABCD) − ½ ar(ABCD) = ½ ar(ABCD) = ar(ΔDPC).
Hence proved: ar(ΔDPC) = ar(ΔAPD) + ar(ΔBPC).
Example 3: Example 3: Using the converse
Problem: ΔABC and ΔDBC are on the same base BC and on the same side of it. ar(ΔABC) = ar(ΔDBC). Prove that AD ∥ BC.
Solution:
- ar(ΔABC) = ½ × BC × h₁ (h₁ = height from A to BC).
- ar(ΔDBC) = ½ × BC × h₂ (h₂ = height from D to BC).
- Given ar(ΔABC) = ar(ΔDBC).
- ½ × BC × h₁ = ½ × BC × h₂
- h₁ = h₂
- A and D are equidistant from line BC, on the same side.
- Therefore, AD ∥ BC.
Hence proved.
Example 4: Example 4: Median divides into equal areas
Problem: In ΔABC, AD is a median. Prove that ar(ΔABD) = ar(ΔACD).
Solution:
Given: AD is a median, so D is the midpoint of BC. BD = DC.
- Let h be the perpendicular distance from A to BC.
- ar(ΔABD) = ½ × BD × h
- ar(ΔACD) = ½ × DC × h
- Since BD = DC (D is midpoint):
- ar(ΔABD) = ar(ΔACD) = ½ × BD × h
Answer: The median divides ΔABC into two triangles of equal area.
Example 5: Example 5: Trapezium diagonal property
Problem: ABCD is a trapezium with AB ∥ DC. Prove that ar(ΔAOB) = ar(ΔDOC) is NOT necessarily true, but ar(ΔADB) = ar(ΔACB).
Solution:
Consider ΔADB and ΔACB:
- Both have the same base AB.
- Vertices D and C both lie on DC, which is parallel to AB.
- Therefore both triangles lie between the same parallels AB and DC.
By the theorem:
- ar(ΔADB) = ar(ΔACB).
Subtracting ar(ΔAOB) from both:
- ar(ΔAOD) = ar(ΔBOC).
Hence proved: In a trapezium, the triangles formed by the diagonals on the non-parallel sides have equal area.
Example 6: Example 6: Area ratio problem
Problem: In ΔABC, D is the midpoint of BC and E is the midpoint of AD. Find the ratio ar(ΔBED) : ar(ΔABC).
Solution:
- AD is a median of ΔABC (D is midpoint of BC).
- ar(ΔABD) = ar(ΔACD) = ½ ar(ΔABC). ... (i)
- In ΔABD, BE is a median (E is midpoint of AD).
- ar(ΔBED) = ½ ar(ΔABD). ... (ii)
- From (i) and (ii): ar(ΔBED) = ½ × ½ × ar(ΔABC) = ¼ ar(ΔABC).
Answer: ar(ΔBED) : ar(ΔABC) = 1 : 4.
Example 7: Example 7: Coordinate geometry application
Problem: A(1, 2), B(5, 2), and C(3, 6) form ΔABC. D(7, 6) is such that ΔDBC has the same base BC. Verify that ar(ΔABC) = ar(ΔDBC).
Solution:
Area of ΔABC:
- = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
- = ½ |1(2 − 6) + 5(6 − 2) + 3(2 − 2)|
- = ½ |−4 + 20 + 0| = ½ × 16 = 8 sq units
Area of ΔDBC:
- = ½ |7(2 − 6) + 5(6 − 6) + 3(6 − 2)|
- = ½ |−28 + 0 + 12| = ½ × 16 = 8 sq units
Verification: Both C(3, 6) and D(7, 6) have y = 6, so CD is horizontal (parallel to line through B extended). The base BC and line through A and D give the same perpendicular distance = 4 units.
Answer: ar(ΔABC) = ar(ΔDBC) = 8 sq units ✓
Example 8: Example 8: Triangle and parallelogram
Problem: A parallelogram ABCD and a triangle ABE are on the same base AB and between the same parallels AB and DC. If ar(ABCD) = 72 sq cm, find ar(ΔABE).
Solution:
Given:
- ABCD is a parallelogram with base AB.
- ΔABE has the same base AB.
- Both lie between parallels AB and DC (vertex E lies on line DC).
Using the theorem:
- ar(ΔABE) = ½ × ar(ABCD) = ½ × 72 = 36 sq cm
Answer: ar(ΔABE) = 36 sq cm.
Example 9: Example 9: Three parallel lines
Problem: Points A and B lie on line ℓ₁. Points C and D lie on line ℓ₂ parallel to ℓ₁. AB = 8 cm and the distance between ℓ₁ and ℓ₂ is 5 cm. Find ar(ΔABC) and ar(ΔABD).
Solution:
- Both ΔABC and ΔABD have the same base AB = 8 cm.
- Vertices C and D lie on ℓ₂, which is parallel to ℓ₁ (containing AB).
- Height of each triangle = distance between parallels = 5 cm.
Area of each:
- ar(ΔABC) = ½ × 8 × 5 = 20 sq cm
- ar(ΔABD) = ½ × 8 × 5 = 20 sq cm
Answer: ar(ΔABC) = ar(ΔABD) = 20 sq cm.
Example 10: Example 10: Finding height from area equality
Problem: ΔPQR has base QR = 12 cm and height from P = 8 cm. ΔSQR has the same base QR and ar(ΔSQR) = ar(ΔPQR). Find the height from S to QR.
Solution:
- ar(ΔPQR) = ½ × 12 × 8 = 48 sq cm
- ar(ΔSQR) = 48 sq cm (given equal)
- ½ × 12 × h = 48
- 6h = 48
- h = 8 cm
Conclusion: S is at the same perpendicular distance (8 cm) from QR as P. Therefore PS ∥ QR.
Answer: Height from S = 8 cm, and PS ∥ QR.
Real-World Applications
Applications:
- Land division: Dividing a triangular plot into equal-area portions by drawing medians or lines parallel to the base.
- Area computation: Finding the area of complex polygons by dividing them into triangles on the same base.
- Proving equal areas: Many geometry proofs require showing two regions have equal area, which this theorem simplifies.
- Parallelogram properties: The theorem that a diagonal bisects a parallelogram into equal areas is a direct corollary.
- Engineering: Load distribution in triangular truss structures uses area relationships.
- Art and design: Creating designs with equal-area triangular regions using parallel guidelines.
Key Points to Remember
- Triangles on the same base and between the same parallels have equal area.
- The key idea: same base + same height = same area.
- The converse is also true: equal areas on the same base implies the vertices lie on a parallel line.
- A median divides a triangle into two triangles of equal area (special case).
- A triangle on the same base and between the same parallels as a parallelogram has half the area of the parallelogram.
- Parallelograms on the same base and between the same parallels are equal in area.
- In a trapezium, ΔAOD = ΔBOC (triangles on either side of the intersection of diagonals).
- Moving the vertex along a line parallel to the base does not change the area.
- This theorem is in Chapter 9 (Areas of Parallelograms and Triangles) of CBSE Class 9.
- All proofs rely on the formula: ar(Δ) = ½ × base × height.
Practice Problems
- Triangles ABC and DBC are on the same base BC, between parallels BC and AD. If BC = 15 cm and the perpendicular distance between the parallels is 8 cm, find the area of each triangle.
- ABCD is a parallelogram. E is any point on DC. Prove that ar(ΔAEB) = ½ ar(ABCD).
- In ΔABC, D is the midpoint of BC. Prove that ar(ΔABD) = ar(ΔACD) = ½ ar(ΔABC).
- ABCD is a trapezium with AB ∥ DC. The diagonals meet at O. Prove that ar(ΔAOD) = ar(ΔBOC).
- In ΔPQR, S and T are points on QR such that QS = ST = TR. Prove that ar(ΔPQS) = ar(ΔPST) = ar(ΔPTR).
- A parallelogram PQRS and triangle PQT are on the same base PQ. If ar(PQRS) = 96 sq cm and T lies on SR, find ar(ΔPQT).
- In ΔABC, medians AD and BE intersect at G. Find ar(ΔBGD) : ar(ΔABC).
- ABCD is a parallelogram. A line through A meets DC at E and BC produced at F. Prove that ar(ΔADF) = ar(ΔAEF).
Frequently Asked Questions
Q1. What does 'between the same parallels' mean?
It means the base of the triangles lies on one parallel line, and the opposite vertices lie on another parallel line. The perpendicular distance between these two parallel lines is the common height.
Q2. Why do triangles on the same base and between the same parallels have equal area?
Because area = ½ × base × height. Both triangles share the same base and the same height (the perpendicular distance between the parallel lines). Therefore their areas are equal.
Q3. Is the converse true?
Yes. If two triangles on the same base have equal areas and lie on the same side of the base, then the line joining their vertices is parallel to the base.
Q4. How is this different from congruent triangles?
Congruent triangles have the same shape and size. Triangles with equal area need not be congruent — they can have completely different shapes. This theorem only guarantees equal areas, not congruence.
Q5. Can this theorem apply to more than two triangles?
Yes. Any number of triangles on the same base, with vertices on the same parallel line, will all have equal area.
Q6. How does this relate to the area of a parallelogram?
A diagonal divides a parallelogram into two triangles on the same base, between the same parallels. By this theorem, both triangles have equal area, each being half the parallelogram's area.
Q7. What if the triangles are on opposite sides of the base?
The theorem still holds. The heights are measured perpendicular to the base. If both heights are equal, the areas are equal regardless of which side the vertex is on.
Q8. Is this in the CBSE Class 9 syllabus?
Yes. This theorem is a key result in Chapter 9 (Areas of Parallelograms and Triangles) of the CBSE Class 9 NCERT Mathematics textbook.
