# Four Isosceles Triangles Hinged at Vertices

### What is this about?

### Problem

Triangles $BAB'$, $CAC',$ $BDC,$ and $BD'C'$ are similar and isosceles, with apex angles at $A,$ $D,$ and $D',$ respectively.

Prove that

$D,$ $A,$ and $D'$ are collinear. (Let $f$ be the line.)

$AD=AD'.$

$BC'$ and $B'C$ meet at point $F$ on $f.$

$BC'=B'C.$

The circumcircles of triangles $ABB',$ $ACC',$ $BCD,$ and $B'C'D'$ concur at $F.$

### Hint

Rotation transform is bound to play an important role for this problem.

### Solution

A product of successive rotations is either a rotation or a translation (which is a rotation through $0^{\circ}.$) In any event, the resulting angle of rotation equals the sum of all the rotations in the sequence.

Consider a sequence of rotations each through the apex angle of the given isosceles triangles:

- Clockwise around $D;$
- Counterclockwise around $A;$
- Clockwise around $D';$
- Counterclockwise around $A;$

Since the angles add up to $0^{\circ},$ the resulting transform is at best a translation. But this translation maps $B$ to itself, implying that in fact it's an identity transform: it leaves the whole plane (point by point) in place. In particular, it leaves in place point $A.$ But the last rotation has $A$ as a fixed point anyway, so that the product of the first three already maps $A$ to itself. It follows that isosceles triangles $ADA'$ and $A''D'A$ are equal and, in particular, $AD=AD',$ proving #2.

... to be continued ...

### Acknowledgment

The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.

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