Read the given facts
Mark midpoints, ratios, parallel lines, known areas, and equal regions.
Chain Equal-Area Transformations
串联等积变形
Advanced geometry problems become manageable when every arrow in the reasoning chain has a reason: midpoint, parallel lines, a shared base or height, a segment ratio, or cancellation of a common area.
One fact. One justified step. One connected chain.
Mission 1
Not completedBuild a proof chain, not a guess
A transformation chain alternates between a fact, a reason, and a new result. The new result becomes the next fact. In area equations, a triangle’s name stands for its area; equal areas do not mean identical shapes.
Given fact
D is a midpoint
D is a midpoint
Reason
equal base halves
equal base halves
New result
one triangle is half
one triangle is half
Next reason
same height ratio
same height ratio
Midpointcreates a 1:1 segment ratio
Parallel linescreate equal perpendicular heights
Same heightturns base ratios into area ratios
Same baseturns height ratios into area ratios
Common areamay be subtracted from equal wholes
Mission 2
Not completedChoose the reason that unlocks the next step
The same diagram may contain many facts. The skill is choosing the one that produces the next useful area relationship.
D is the midpoint of BC. The whole triangle has area 24 square units.
1
Segment fact: BD = DC, so BD is one half of BC.
2
Area reason: triangles ABD and ABC use bases BD and BC on the same line and share the height from A.
3
Result: Area(ABD) = ½ × Area(ABC) = 12.
midpoint → base ratio 1:2 → area ratio 1:2
Mission 3
Not completedA moving point, a fixed area
Triangle ABC has area 24 square units and M is the midpoint of AB. E moves along AM, and D lies on BC. Before E reaches M, D is chosen so that MD is parallel to EC. Even though the diagram changes, Area(EBD) stays fixed.
Move E from A toward M. Point D moves automatically so that MD remains parallel to EC. The drawing is not necessarily to scale.
Whole area24
Target area12
BD as fraction of BC0.61
E-height as fraction of A-height0.82
Why the product stays fixed: as E moves downward, its height decreases, but D moves right so BD increases. The equal-area proof below explains why they balance.
Proof — why the area stays 12
- Join M to C. Triangles EMD and CMD share base MD. Their third vertices E and C lie on EC, parallel to MD, so the triangles have equal areas.
- Add triangle MBD to both. EMD + MBD makes EBD; CMD + MBD makes MBC. Therefore Area(EBD) = Area(MBC).
- M is the midpoint of AB. Triangles MBC and ABC share the height from C to AB, and MB is half of AB. So Area(MBC) = 24 ÷ 2 = 12.
At the final slider position, E coincides with M and D with C. The two lines then coincide rather than being distinct parallel lines, and EBD is simply MBC, still of area 12. The small comparison triangles have collapsed to zero area.
Mission 4
Not completedChain two segment ratios through one triangle
In triangle ABC, E lies between B and C and D lies between A and C. Given BC = 3BE and AC = 4CD. The target is the ratio Area(ABC) : Area(DEA).
1
BC = 3BE, so BE/BC = 1/3 and EC/BC = 2/3.
2
Triangles AEC and ABC share the height from A to BC, so Area(AEC) = 2/3 Area(ABC).
3
AC = 4CD, so CD/AC = 1/4 and AD/AC = 3/4.
4
Triangles AED and AEC share the height from E to AC, so Area(AED) = 3/4 Area(AEC).
5
3/4 × 2/3 = 1/2, so Area(ABC) is twice Area(DEA).
Mission 5
Not completedUse equal areas to reveal a hidden parallel line
In triangle ABC, D lies between A and B, E lies between B and C, and F lies between A and C. AD = 2 units, DB = 3 units, and quadrilateral DBEF has the same area as triangle ABE. The surprising goal is to find that common area when Area(ABC) = 10.
1
Quadrilateral DBEF = triangle DBE + triangle DEF.
2
Triangle ABE = triangle DBE + triangle ADE.
3
The two wholes are equal, so subtract common triangle DBE: Area(DEF) = Area(ADE).
4
Triangles DEF and ADE share base DE. Their distinct third vertices F and A lie on the same side of DE. Equal areas with the same base mean equal perpendicular distances to DE. Points on that side at a fixed distance lie on one parallel line. Since A and F lie on AC, DE ∥ AC.
5
Triangles ACD and ACE share base AC, and D and E lie on the parallel line DE. Their areas are equal. Subtract each from ABC: Area(BCD) = Area(ABE).
6
Triangles BCD and BCA share the height from C to AB. BD/BA = 3/5, so Area(BCD) = 3/5 × 10 = 6. Thus ABE and DBEF each have area 6. Also, ABE and ABC share the height from A to BC, so BE/BC = 6/10 = 3/5.
Mission 6
Not completedSend one ratio around a trapezoid network
In trapezoid ABCD, AD is parallel to BC and diagonals AC and BD meet at E. Area(ABE) = 30 square units and EC = 2AE. The diagonal network lets one 1:2 ratio determine all four small triangles.
| Region | Reason | Area |
|---|---|---|
| ABE | given | 30 |
| CDE | opposite side triangles in a trapezoid | 30 |
| BCE | EC = 2AE, same height to AC | 60 |
| ADE | AE:EC = 1:2, same height to AC | 15 |
30 + 30 + 60 + 15 = 135
Mission 7
Not completedNest a midpoint chain inside a segment-ratio chain
Triangle ABC has area 72 square units. D is the midpoint of BC, E lies between A and B with BE = 3AE, and F lies between E and D with FD = 2EF. Find Area(AFD).
1
D is the midpoint of BC, so Area(ABD) = 72 ÷ 2 = 36.
2
BE = 3AE, so AE/AB = 1/4. With the same height from D, Area(AED) = 36 ÷ 4 = 9.
3
FD = 2EF, so FD/ED = 2/3. Triangles AFD and AED share the height from A to line ED.
4
Area(AFD) = 2/3 × 9 = 6.
Mission 8
Not completedComplete a rectangle chain—and debug weak links
Rectangle ABCD has AD = 10 cm and DC = 6 cm. E is the midpoint of AD, F is the midpoint of DC, and G lies between E and F with EG = 2GF. Find Area(BFG).
1
Rectangle area: 10 × 6 = 60.
2
AE = ED = 5 and DF = FC = 3. Triangle ABE has area 5×6÷2 = 15; triangle BCF has area 10×3÷2 = 15; triangle EDF has area 5×3÷2 = 7.5. These are the three nonoverlapping corners outside BEF.
3
So Area(BEF) = 60 − 15 − 15 − 7.5 = 22.5.
4
EG = 2GF, so GF/EF = 1/3. Triangles BFG and BFE share the height from B to line EF.
5
Area(BFG) = 1/3 × 22.5 = 7.5.
Extra practice — put the relationships together
These additional problems combine the relationships used in this lesson. Try these two diagrams before the workshop. Each extra practice has its own checkpoint in the lesson navigation.
Prove the fraction made by three midpoints
original Exercise 4
In parallelogram BCDE, M and N are the midpoints of BE and CD. A is the midpoint of MN. Triangle BAN is shaded and has area 15 cm². Find the area of the whole parallelogram.
Use the stated relationships; do not measure the drawing.
Try each step first. Open a hint if you need a starting point.
Hint — choose the first relationship
First compare BAN and BMN: their bases AN and MN lie on the same line, and they share the height from B.
Worked solution — compare after trying
- Since A is the midpoint of MN, AN is half of MN. Therefore BAN is half of BMN, so BMN = 2 × 15 = 30.
- M and N are halfway along the parallel sides. MN is parallel and equal to BC, halfway up the whole parallelogram. BCNM is the lower half.
- Diagonal BN halves parallelogram BCNM. Thus BCNM = 2 × BMN = 60.
- The whole is twice this lower half: 2 × 60 = 120 cm². The shaded triangle is ½ × ½ × ½ = ⅛ of the whole.
Track both shaded pieces in an intersection problem
original Exercise 11
Triangle ABC has area 14 cm². D lies on BC with DC = 3DB. E is the midpoint of AD. Line CE meets AB at F. Find the combined area of shaded triangles AFE and DEC.
Use the stated relationships; do not measure the drawing.
Try each step first. Open a hint if you need a starting point.
Hint — choose the first relationship
DC is three quarters of BC, so ADC is three quarters of ABC. Then halve ADC using midpoint E. To find the other piece, give Area(AFE) the temporary name x.
Worked solution — compare after trying
- DC/BC = 3/4. With the same height from A, ADC = 14 × 3/4 = 10.5.
- AE = ED, so AEC and DEC have equal bases on AD and share the height from C. Each has area 10.5 ÷ 2 = 5.25.
- Let Area(AFE) = x. Since AE = ED, AFE and DFE share the height from F to AD and have equal areas. Thus AFD = 2x.
- Now compare AFD and AFC, using the common base AF on line AB. On ray BC, BC = 4BD; the perpendicular distance from C to AB is four times the distance from D to AB. So AFC = 4 × AFD = 8x.
- Because F, E, C are collinear, AFC also consists of AFE and AEC: its area is x + 5.25. Therefore 8x = x + 5.25, so 7x = 5.25 and x = 0.75.
- Add both shaded pieces: AFE + DEC = 0.75 + 5.25 = 6 cm². Finding DEC alone does not finish the practice problem.
Mission 9
Not completedReasoning-chain workshop
Solve at least six of eight. Use the named mission or linked practice for the point arrangement. Give areas in square units unless cm² is stated. Before entering a number, follow the chain: fact → reason → result.
Triangle ABC has area 30 and D is the midpoint of BC. Find Area(ABD).
In a parallelogram, M and N are side midpoints and A is the midpoint of MN. Shaded triangle BAN has area 15. Use the three-midpoint diagram to find the parallelogram area.
Use Mission 4’s point arrangement, with BC=3BE and AC=4CD. Area(ABC) is how many times Area(DEA)?
Use Mission 5’s point arrangement: AD=2, DB=3, Area(ABC)=10, and quadrilateral DBEF has the same area as triangle ABE. Find their common area.
In Mission 6’s trapezoid, Area(ABE)=30 and EC=2AE. Find the trapezoid area.
Use Mission 7’s point arrangement: Area(ABC)=72, D is the midpoint of BC, BE=3AE, and FD=2EF. Find Area(AFD).
Area(ABC)=14, DC=3DB, E is the midpoint of AD, and C, E, F are collinear with F on AB. Find the total shaded area AFE + DEC using the two-shaded-region diagram.
In Mission 8’s rectangle, E and F are the stated midpoints and G lies on EF with EG=2GF. Find Area(BFG).
Workshop hint — choose your first step
Draw or revisit the diagram. Write one arrow for each midpoint, shared height, or equal remainder; explain the reason before calculating.
Worked workshop solutions — open after trying
Compare your first different step, then close this panel and try again.
- The midpoint halves the base and the triangles share a height: ABD = 30 ÷ 2 = 15.
- BAN is half of BMN, BMN is half of lower parallelogram BCNM, and BCNM is half of the whole. Reverse the three halvings: 15 × 2 × 2 × 2 = 120.
- EC/BC = 2/3 and AD/AC = 3/4. Shared-height comparisons give DEA/ABC = (2/3) × (3/4) = 1/2, so ABC is twice DEA.
- Subtract DBE to obtain equal DEF and ADE, proving DE ∥ AC. ACD and ACE then have equal areas. Subtract them from ABC: BCD = ABE. BCD = (3/5) × 10 = 6, the common area.
- CDE = ABE = 30. AE : EC = 1 : 2 gives BCE = 60 and ADE = 15. Whole = 30 + 30 + 60 + 15 = 135.
- Midpoint gives ABD = 36. AE/AB = 1/4 gives AED = 9. FD/ED = 2/3 gives AFD = 6.
- The worked example practice derives DEC = 5.25. If AFE = x, midpoint E gives AFD = 2x; the height ratio gives AFC = 8x. Also AFC = x + 5.25, so x = 0.75. Both shaded pieces total 6.
- Rectangle = 60. Remove corner areas 15, 15, and 7.5 to get BEF = 22.5. GF/EF = 1/3, so BFG = 22.5 ÷ 3 = 7.5.
Mission 10
Not completedExit ticket
Answer all five questions. Every scored answer is objectively checked.
Teaching notes
The proof-chain organizer, movable diagram, original-independent explanations, diagnostic questions, and additional workshop problems are instructional scaffolds created for this page. All diagrams are redrawn and are not necessarily to scale.