Why triangles between the same parallels always have equal area — and how to construct equivalent figures.
Parallelogram TheoremSame Base, Same ParallelsMedian Area TheoremArea Constructions8 Board Questions
Core Concepts
Parallelogram Area Theorem
Two parallelograms that share the same base and lie between the same pair of parallel lines have equal areas. This is because they have the same base length and the same perpendicular height (the distance between the parallel lines).
Picture this: Imagine a deck of cards. Push the deck sideways — you get a leaning parallelogram. The area of the cross-section stays the same as the upright rectangle, because the base and height don't change.
Triangle and Parallelogram (Same Base)
If a triangle and a parallelogram share the same base and lie between the same parallel lines, then:
Area of Triangle = ½ × Area of Parallelogram
This follows directly from the fact that a diagonal divides a parallelogram into two equal triangles.
Triangles on Same Base Between Same Parallels
Any two triangles that share the same base (or equal bases) and have their apex on the same line parallel to that base have equal areas. Moving the apex horizontally along the parallel line doesn't change the base or height.
Median Divides Triangle into Equal Areas
A median is a line from a vertex to the midpoint of the opposite side. Every median divides the triangle into two smaller triangles of exactly equal area. This is because both triangles share the same height from the vertex, and their bases (the two halves of the divided side) are equal.
Area Division by a Cevian
If a point D divides BC in ratio m:n, then: Area(△ABD) / Area(△ACD) = m/n. This is because both triangles share the same height from A, and their bases BD and DC are in ratio m:n.
Construction: Triangle Equal in Area to Quadrilateral
To construct a triangle with the same area as quadrilateral ABCD: Draw diagonal AC. Through D, draw a line parallel to AC meeting BC extended at E. Then △ABE has the same area as quadrilateral ABCD. (△ADE = △ACE since they share base AE and are between same parallels, so removing △ACE from the quad and replacing with △ADE gives the same area.)
Key Theorems
Triangle Area
A = ½ × base × height
Height must be perpendicular to the base
Parallelogram Area
A = base × height
Height = perpendicular distance between parallel sides
Triangle = ½ Parallelogram
A(△) = ½ · A(▱) (same base, same parallels)
Cevian Area Division
A(△ABD) / A(△ACD) = BD / DC
D divides BC; both triangles share height from A
Median
A(△ABM) = A(△ACM) = ½ · A(△ABC)
M is midpoint of BC
Area Preservation Visualiser
Drag the apex point horizontally. The area stays constant because the base and height are unchanged — demonstrating the "same parallels" theorem.
Interactive Area Invariant Demo
Drag the apex left or right — notice the area never changes!
Board Questions
Q1Dhaka 2023Parallelogram = Rectangle Area5 Marks
Parallelogram ABCD and rectangle ABEF are on the same base AB and between the same parallels AB ∥ EF (where EF passes through D and C). Prove that they have equal area.
1
Given: ABCD is a parallelogram and ABEF is a rectangle, both on base AB and between parallels AB and DC (= EF line).
2
In △ADF and △BCE: AD = BC (opp. sides of parallelogram), AE = BF (opp. sides of rectangle), ∠DAF = ∠CBE = 90° (rectangle has right angles).
3
Wait — more directly: Both figures share the same base AB and lie between the same parallel lines AB and EF (= DC extended). By the parallelogram area theorem, Area(ABCD) = base × height = AB × h.
4
Area of rectangle ABEF = AB × h (where h = perpendicular distance between AB and EF). Both equal AB × h.
Ans
Area(ABCD) = Area(ABEF) = AB × h ✓
Q2Rajshahi 2024Median Theorem5 Marks
In △ABC, D is the midpoint of BC. Prove that Area(△ABD) = Area(△ACD).
1
Given: D is midpoint of BC, so BD = DC. AD is the median.
2
Draw AE ⊥ BC (height from A). Let AE = h. This height is common to both △ABD and △ACD.
3
Area(△ABD) = ½ × BD × h.
4
Area(△ACD) = ½ × DC × h.
5
Since D is midpoint, BD = DC. Therefore Area(△ABD) = Area(△ACD). QED ✓
Q3Chittagong 2022Triangle = ½ Parallelogram5 Marks
ABCD is a parallelogram. E is any point on CD. Prove that Area(△ABE) = ½ × Area(parallelogram ABCD).
1
△ABE has base AB (same as parallelogram ABCD). E lies on CD, which is parallel to AB.
2
The height of △ABE from E to AB equals the perpendicular distance h between AB and CD (the parallel sides of the parallelogram).
3
Area(△ABE) = ½ × AB × h.
4
Area(ABCD) = AB × h.
5
Therefore Area(△ABE) = ½ × Area(ABCD). This holds for any position of E on CD. QED ✓
Q4Jessore 2021Median Proof (Detailed)6 Marks
In △ABC, median AM is drawn to BC. Prove that the median divides the triangle into two triangles of equal area.
1
Given: △ABC with AM as median, so M is midpoint of BC → BM = MC.
2
To prove: Area(△ABM) = Area(△ACM) = ½ Area(△ABC).
3
Construction: Draw AN ⊥ BC, the altitude from A. Let AN = h.
4
Area(△ABM) = ½ × BM × h (base BM, height h).
5
Area(△ACM) = ½ × MC × h (base MC, same height h from A).
6
Since M is midpoint, BM = MC. Therefore Area(△ABM) = Area(△ACM).
7
Also Area(△ABM) + Area(△ACM) = Area(△ABC), so each = ½ × Area(△ABC). QED ✓
Q5Comilla 2023Area Ratio — Parallelogram and Triangle5 Marks
Parallelogram PQRS and triangle PQT share the same base PQ. T lies between lines PQ and SR. Find the ratio of areas Area(PQRS) : Area(△PQT).
1
Both PQRS and △PQT lie between parallels PQ and SR (since T is between them).
2
Let PQ = b and perpendicular distance between parallels = h.
3
Area(PQRS) = b × h.
4
Area(△PQT) = ½ × b × h' where h' is the distance from T to PQ. Since T is between the parallels, h' ≤ h.
5
If T lies exactly on SR, then h' = h and Area(△PQT) = ½bh. In this case Area(PQRS) : Area(△PQT) = bh : ½bh = 2 : 1.
Ans
Area(PQRS) : Area(△PQT) = 2 : 1 (when T is on line SR) ✓
Q6Barisal 2024Construction — Triangle = Quadrilateral8 Marks
Construct a triangle equal in area to quadrilateral ABCD where AB = 5 cm, BC = 4 cm, CD = 3 cm, DA = 6 cm, diagonal AC = 7 cm.
1
Method: Draw the quadrilateral ABCD using the given measurements. Draw diagonal AC.
2
Step 2: Through vertex D, draw a line DE parallel to diagonal AC (using ruler and compass: draw a line through D parallel to AC).
3
Step 3: Extend side BC to meet this parallel line at point E.
4
Why it works: △ACD and △ACE have the same base AC and lie between the same parallels AC and DE, so Area(△ACD) = Area(△ACE).
△ABE is the required triangle with Area(△ABE) = Area(ABCD) ✓
Q7Sylhet 2022Equal Areas — Height Relationship4 Marks
Two triangles have equal areas and share the same base but have different heights. What can you conclude about their heights? Explain.
1
Let both triangles have base b. Let heights be h₁ and h₂.
2
Area₁ = ½ × b × h₁ and Area₂ = ½ × b × h₂.
3
If Area₁ = Area₂, then ½bh₁ = ½bh₂ → h₁ = h₂.
4
Conclusion: If two triangles on the same base have equal areas, their heights must be equal. This means their apex points lie on the same line parallel to the base (the "same parallels" condition).
Ans
Their heights must be equal — apex points lie on a line parallel to the base ✓
Q8Dinajpur 2023Cevian Area Division5 Marks
△ABC has area 48 cm². D divides BC in ratio 1:2. Find the area of △ABD.
1
D divides BC such that BD:DC = 1:2. So BD = (1/3) × BC and DC = (2/3) × BC.
2
△ABD and △ABC share the same height from vertex A (let it be h).