Chapter 8.4 — Exercise 8.3 — Areas of Similar Triangles

Deriving the Area-Ratio Theorem

Given △ABC ~ △PQR, drop altitudes AM and PN onto BC and QR. Since both triangles share the base-times-height area formula:

△ABM ~ △PQN by AA (∠B=∠Q from the original similarity, ∠M=∠N=90°), which gives AM/PN = AB/PQ — the altitude ratio and the side ratio turn out to be identical, which is the single fact this whole derivation hinges on. Since △ABC~△PQR also directly gives AB/PQ = BC/QR, substituting both facts back into the area ratio above turns BC/QR × AM/PN into BC/QR × AB/PQ — and because those two ratios are equal, the whole expression collapses into a single ratio squared. Nothing here needed the triangles' actual angle measures or side lengths — only that a matching pair of altitudes forms its own pair of similar right triangles, a fact that follows automatically from the original similarity rather than needing to be assumed separately.

Any one of the three side pairs gives the identical area ratio, since all three were already equal to each other by the original similarity — there's no need to check all three sides separately once similarity is already established; whichever one length ratio a problem happens to supply is enough on its own.

The Midpoint Triangle Is Always a Quarter

D, E, F are the midpoints of BC, CA, AB in △ABC. Since AF/FB=1 and AE/EC=1, the converse of BPT gives FE∥BC — and the same reasoning on the other two sides makes BDEF a parallelogram, forcing FE=BD, so FE/BC=½. The identical argument applies to the other two sides, giving DE/AB=½ and DF/AC=½ as well.

The triangle formed by joining a triangle's three midpoints is always exactly a quarter of the original's area — a fact worth remembering on its own, since it holds for every triangle regardless of shape, not just specific ones with tidy coordinates. It's also a genuinely quick way to spot a 1:4 area ratio without touching the general formula at all: the moment a problem mentions three midpoints joined together, the answer is already known before a single length is measured.

Bisecting Area, Not Just Length

In △ABC, XY is drawn parallel to AC (X on AB, Y on BC), splitting the triangle into two equal-area pieces: △BXY and the remaining quadrilateral AXYC. Finding AX/XB from that single area condition takes one extra step beyond the theorem itself.

Since △XBY~△ABC (AA, sharing ∠B with XY∥AC), the area-ratio theorem converts that "2" directly into a side ratio: (AB/XB)² = 2, so AB/XB = √2. Since AX = AB−XB, dividing through by XB gives AX/XB = √2 − 1 — an irrational ratio, which is worth expecting here since nothing in the problem's setup guarantees a clean whole-number split. It's worth double-checking the subtraction step specifically: AB/XB − 1 equals (AB−XB)/XB only because AB/XB and 1 share the same denominator XB once 1 is rewritten as XB/XB — a small algebraic move that's easy to skip past without noticing why it's valid.

The Same Squared Ratio, for Medians and Altitudes

The area-ratio theorem isn't limited to full sides — any pair of corresponding lengths in similar triangles obeys the identical squared relationship, including medians and altitudes.

• Medians: if CM and RN are corresponding medians of △ABC~△PQR, then AM/PN = AC/PR (SAS similarity, using that M and N are midpoints), which gives CM/RN = AC/PR directly — and combined with the area-ratio theorem, ar(△ABC)/ar(△PQR) = (CM/RN)² exactly as it would with any full side. This proof leans directly on the same half-side median fact used earlier in Exercise 8.2 to prove two triangles' medians similar in the first place — this exercise simply carries that result one step further, from a side-length ratio into a squared area ratio.
• Altitudes: given two similar triangles with areas 81 cm² and 49 cm², and the larger one's altitude measuring 4.5 cm, the same squared relationship runs in reverse to find the smaller altitude: 81/49 = (4.5/x)², so 4.5/x = 9/7, giving x = 3.5 cm.

Sides, medians, altitudes — whichever corresponding length a problem hands over, the same square-the-ratio relationship applies without any adjustment. The altitude problem above is worth a second glance for what it does differently from every other problem in this exercise: it starts from a known area ratio and works backward to an unknown length, rather than starting from a known length ratio and working forward to an area — the same theorem, just run in reverse.

Two Direct Applications

The second row is worth noticing for what it proves before it even reaches the area ratio: PQ∥BC isn't handed over as a given fact, it has to be established first from the equal ratios AP/PB=AQ/QC, using the converse of BPT — the same theorem this entire chapter opened with, still doing work three exercises later. The first row, by contrast, starts from similarity already given outright, which makes it the more direct of the two: no proof step is needed before the area-ratio formula can be applied, just a straightforward substitution of the known side lengths and area.

Both problems are also a useful reminder of which direction the theorem runs in each case. The first finds an unknown area from two known side lengths and one known area — a forward application. The second finds a fraction of one triangle's area purely from two segment ratios, without either triangle's actual area ever being stated as a number — the theorem here isn't computing an area at all, only a relationship between two areas expressed as a fraction.

From Squared Ratios to a Squared Sum

Every result in this exercise came from squaring a single ratio between two similar triangles. Exercise 8.4 uses the same similar-triangles machinery — an altitude splitting a right triangle into two smaller triangles, both similar to the original — to derive a squared relationship of a different kind entirely: the Pythagoras theorem itself. It's worth noticing the family resemblance before starting that exercise: both this exercise and the next build everything from an altitude dropped inside a triangle, creating a pair of smaller similar triangles whose ratios then get squared to reach the final result. For the AA/SSS/SAS criteria this exercise's every similarity claim depends on, revisit Exercise 8.2.