### The Fundamental Theorem of Directly Similar Figures

First, "directly similar" means similar in the same rotational sense, as
opposed to "inversely similar", which means similar in the opposite rotational
sense.

The FTDSF says if the lines connecting the corresponding vertices of two
directly similar polygons are divided in equal ratios, then the resulting
polygon is directly similar to the given two polygons.

Expressed in set notation, the theorem is: Let F_{0} and F_{1}
denote two directly similar figures in the plane, where P_{1}
F_{1} corresponds to P_{0}
F_{0} under the given similarity. Let r
(0,1), and define F_{r} = {(1-r)P_{0}+rP_{1} : P_{0}
F_{0},
P_{1}
F_{1}}. Then F_{r} is also directly similar to F_{0}.

By considering the vertices of the first square in order BCDA, and the second
in order PQRS, with r=1/2, then the midpoints of BP, CQ, DR, and AS are the
vertices of the red figure, which by FTDSF is similar to the other squares.

### Proof

proof based on post by Tom L, . . . . . . needs to be fixed to change "k" to
"r", and add a diagram.

('Euclidean' proof)

It's fairly clear that the result will follow from the following:

Let AB, CD be segments and P be a point ('centre of spiral similarity') such
that ABP, CDP are similar. Then for E,F with kAE=(1-k)EC, kBF=(1-k)FD, EFP is
similar to the first two triangles.

Proof (drawing pictures will make it obvious what I'm doing):

Firstly, it's a nice trivial angle chase (or a 'standard result on spiral
similarities' if you like) to show that ACP and BDP are similar.

Now, our funny k-ratio condition really comes down to the fact that PBF is
similar to PAE (since PAC similar to PBD and E,F are points on the sides AC,BD
with the same ratio).

But now we are basically home... as <FPE = <FPA-<EPA = <FPA-<FPB = <BPA, and
FP/EP = BP/AP. So PAB is similar to PEF.

I'll leave the details on why this implies the whole result to you guys...

### Internet references

Cut-the-knot:
Similar Triangles, which explains the Fundamental Theorem of Directly
Similar Figures

### Related pages in this website:

Summary of geometrical theorems

The Finsler-Hadwiger
Theorem -- Squares ABCD and PQRS joined at a corner D=P; the figure formed
by joining the midpoints of AC, CQ, QS, and SA is a square. This is a
special case of the FTDSF.

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Graeme McRae.