Similar triangles and homothety: lengths times k, area times k²

Similarity & homothety

When a triangle is scaled from a point, why do its angles stay put, and why does the area grow by k² rather than k?

Pick a centre O and a ratio k. Every vertex slides along its ray from O to k times the distance. The board measures the new sides, angles and area: each side ratio is |k|, each angle is unchanged and the area ratio is k².

Drag the ringed points on the board.

Middle school
High school
Homothety (dilation)
Similar triangles
Ratio of similitude
Area ratio k²

k

congruent

point symmetry

2.00

O_x

-5.0

O_y

-2.0

A_x

-2.0

A_y

1.0

\dfrac{S'}{S} = k^2

$|A'B'| / |AB|$: 2.000
$|B'C'| / |BC|$: 2.000
$|C'A'| / |CA|$: 2.000
$\angle A$: 71.6
$\angle A'$: 71.6
$S_{A'B'C'} / S_{ABC}$: 4.000
$k^2$: 4.000

A'B'C' issame way round

Guided tour

1Double every distance from O
With $k=2$ , each vertex moves along its ray from $O$ to twice the distance: $A'=O+2(A-O)$ . The board measures all three new sides, and each ratio comes out $2$ . The angles at $A$ and at $A'$ are both $71.6^\circ$ .
2Area grows by k²
The lengths doubled, but the area ratio reads $4$ , not $2$ : base and height both doubled. At $k=1.5$ the sides grow by $1.5$ and the area by $2.25$ .
3The centre can be anywhere
Drag $O$ inside the triangle, or far away. The image moves, but the side ratios stay $1.5$ and the angles stay equal: where $O$ sits only decides where the copy lands, not its shape.
4Negative k: through the centre
At $k=-1$ each vertex goes through $O$ to the same distance on the other side. The copy is turned half a revolution but is congruent: ratios $1$ , area ratio $1$ . Any negative $k$ does the same with scaling $|k|$ .

Explanation

1
Homothety
Fix a point $O$ and a number $k\ne0$ . The homothety with centre $O$ and ratio $k$ sends each point $X$ to
$X'=O+k\,(X-O)$
so $X'$ lies on line $OX$ and $|OX'|=|k|\,|OX|$ . For $k<0$ it lands on the other side of $O$ .
2
Why every side scales by |k|
Subtract the definition for two points:
$A'-B'=\big(O+k(A-O)\big)-\big(O+k(B-O)\big)=k\,(A-B)$
$O$ cancels. So $A'B'$ is parallel to $AB$ and $|k|$ times as long, and the same holds for every pair of points. This is the whole reason the side ratios agree and do not depend on where $O$ is.
3
Why angles stay the same
Each side of the image is parallel to the matching side of the original. The angle at $A'$ is made by two lines parallel to the two lines at $A$ , and two pairs of parallel lines meet at equal angles (corresponding angles). So the triangles are similar: equal angles, proportional sides.
The converse is what makes the idea general: any two similar triangles can be moved by a rigid motion until they are in homothetic position, so everything proved here holds for all similar figures.
4
Why area scales by k²
Area is $\tfrac12\cdot\text{base}\cdot\text{height}$ . The base is a length, so it scales by $|k|$ . The height is the distance from a vertex to the opposite side, also a length, so it scales by $|k|$ too. The product scales by $k^2$ :
$\frac{S'}{S}=\frac{\tfrac12\,(|k|\,b)(|k|\,h)}{\tfrac12\,bh}=k^2$
For the same reason volume scales by $|k|^3$ : three lengths multiply.

Check yourself

Q01

Two similar triangles have sides in ratio $3:5$ . What is the ratio of their areas?

Q02

Under a homothety with centre $O$ , why is $A'B'$ parallel to $AB$ ?

Q03

A map is drawn at scale $1:1000$ . A field covers $12\,\text{cm}^2$ on the map. How large is it in reality?

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