Chapter 11.4 — Exercise 11.3 — Triangles
Triangles on the same base and between the same parallels. This is Lesson 4 of 4 in Chapter 11: Areas.
Triangles Compared Directly, No Parallelogram Needed
Two triangles sitting on the same base — or on equal bases — between the same parallels are always equal in area. This chapter's closing exercise applies that single fact, together with the median-bisects-area rule from Exercise 11.1, across six genuinely different figures — each one built by chaining these two rules two or three times in a row, rather than needing any new theorem introduced along the way.
Why a Median Keeps Reappearing
A median splits any triangle into two triangles of equal area, since both halves share the same height and equal bases by definition. This one fact, first used back in Exercise 11.1, turns out to be the real engine behind nearly every proof in this exercise — chained two or three times in sequence, it reaches results that look far more complicated than "equal bases, equal heights" on first glance. Spotting a hidden median inside a larger figure — a segment that isn't labeled as a median outright, but happens to connect a vertex to the midpoint of the opposite side purely because some other given fact forces that midpoint condition — is really the one recurring skill every problem below is quietly testing.
A Median's Midpoint Creates a Quarter-Area Triangle
In △ABC, AD is a median, and E is the midpoint of AD; prove area of △ABE = area of △ACE = ¼ area of △ABC. Since AD is a median of △ABC, area of △ABD = area of △ACD = ½ area of △ABC. Since E is the midpoint of AD, BE is itself a median of △ABD, so area of △ABE = ½ area of △ABD = ½ × ½ area of △ABC = ¼ area of △ABC. The identical reasoning applied to △ACD, using CE as its median, gives area of △ACE = ¼ area of △ABC too — so area of △ABE = area of △ACE, both equal to a quarter of the original triangle.
AD median of △ABC → ar(△ABD) = ar(△ACD) = ½ ar(△ABC)
BE median of △ABD → ar(△ABE) = ½ ar(△ABD) = ¼ ar(△ABC)This is the median-bisects-area rule applied twice in a row, once at the level of the whole triangle and once again inside one of the two halves it produces — each application halves the area again, which is exactly why the final answer lands on a quarter rather than a half. The same nested-median idea could, in principle, be pushed one level further still — taking the midpoint of BE itself would carve out an eighth of the original triangle's area, and so on indefinitely, each further nesting halving the previous fraction exactly the same way the first two steps did here.
A Parallelogram's Diagonals Chain Four Medians Together
Show that the diagonals of a parallelogram divide it into four triangles of equal area. Since a parallelogram's diagonals bisect each other at O, AO is a median of △ABD (giving area △AOD = area △AOB), BO is a median of △ABC (giving area △AOB = area △BOC), CO is a median of △BCD (giving area △BOC = area △COD), and DO is a median of △ACD (giving area △COD = area △AOD). Chaining all four equalities together: area △AOD = area △AOB = area △BOC = area △COD.
Four separate applications of the identical median rule, each one borrowed from a different triangle hiding inside the same parallelogram, is what closes this chain into a single loop — the last equality (△COD = △AOD) connects back to the very first triangle named, confirming all four areas match each other rather than merely showing they're equal in scattered, overlapping pairs along the way.
A Bisected Segment Forces Equal Triangle Areas
△ABC and △ABD share base AB; segment CD is bisected by AB at point O; prove area of △ABC = area of △ABD. In △ADC, AO is a median (O bisects CD), so area of △AOC = area of △AOD. In △BDC, BO is a median for the same reason, so area of △BOC = area of △BOD. Adding these two equalities: area of △AOC + area of △BOC = area of △AOD + area of △BOD, which is exactly area of △ABC = area of △ABD.
Point O isn't a vertex of either triangle being compared — it sits on segment CD, entirely separate from the shared base AB — which makes this proof structurally different from a simple same-base-same-parallels argument: the equal areas here come from two medians in two different smaller triangles, added together, rather than from one single direct area comparison made between the two full triangles at once. Finding this hidden pair of medians requires noticing that O, defined here only as the bisection point of segment CD, quietly plays the role of a median's midpoint inside two entirely separate triangles simultaneously — △ADC and △BDC — even though neither of those two triangles was mentioned anywhere in the original problem statement itself.
Turning the Area Theorem Around
D and E lie on sides AB and AC of △ABC, with area of △DBC = area of △EBC; prove DE ∥ BC. Both triangles share base BC, so equal areas force equal heights: calling the perpendicular distances from D and E to BC as h₁ and h₂, ½ × BC × h₁ = ½ × BC × h₂ gives h₁ = h₂ directly. Equal perpendicular distances from D and E onto the same line BC means D and E are equally far from BC — which is exactly the condition for DE to run parallel to BC.
This problem runs the entire chapter's central theorem in reverse: instead of starting from "same base, same parallels" and concluding "equal area," it starts from equal area on a shared base and concludes the two points must lie on a line parallel to that base — exactly the same underlying relationship, simply read carefully in the opposite direction this time.
Diagonals of a Trapezium, Not a Parallelogram
Trapezium ABCD has AB ∥ DC, with diagonals AC and BD meeting at O; prove area of △AOD = area of △BOC. Since △ADB and △ACB share base AB and sit between the same parallels AB and DC, area of △ADB = area of △ACB. Subtracting the shared triangle △AOB from both sides: area of △ADB − area of △AOB = area of △ACB − area of △AOB, which simplifies directly to area of △AOD = area of △BOC.
Exercise 11.1 proved this identical-looking result for a parallelogram using SSS congruence between the two triangles directly. Here, with only one pair of sides parallel rather than two, △AOD and △BOC generally aren't congruent at all — they can have completely different shapes — yet the "subtract a shared piece from two equal wholes" argument still forces their areas to match exactly, a genuinely different route to an equal-area conclusion than congruence provides.
Six Proofs, One Recurring Engine
| Figure | Core technique |
|---|---|
| Median's midpoint E | Median rule applied twice → quarter area |
| Parallelogram diagonals | Median rule applied four times in a loop |
| Bisected segment CD | Two medians in two triangles, added together |
| Equal-area points D, E | Same-base theorem run in reverse |
| Trapezium diagonals | Subtracting a shared triangle from two equal wholes |
Straight Lines and Angles, Traded for Curves
Every figure across this entire chapter — parallelograms, triangles, trapeziums, rhombuses — was built from straight sides meeting at sharp vertices. Chapter 12, Circles leaves straight edges behind almost entirely, introducing a shape with no vertices, no sides, and no straight lines anywhere in its boundary at all — a genuinely different kind of geometry, starting fresh from a single entirely new definition.