Double perspective triangles

Double perspective triangles

Two triangles in double perspective are in triple perspective

Exeter B.png
Exeter C.png

Let $\triangle ABC$ and $\triangle DEF$ be in double perspective, which means that triples of lines $AF, BD, CE$ and $AD, BE, CF$ are concurrent. Prove that lines $AE, BF,$ and $CD$ are concurrent (the triangles are in triple perspective).

Proof

Denote $G = AF \cap BE.$

It is known that there is projective transformation that maps any quadrungle into square.

We use this transformation for $BDFG$. We use the Claim for square and get the result: lines $AE, BF,$ and $CD$ are concurrent.

Claim for square

Let $ADBG$ be the square, let $CEGF$ be the rectangle, $A \in FG, G \in BE.$

Prove that lines $BF, CD,$ and $AE$ are concurrent.

Proof

Let $BG = a, GE = b, AF = c, A = (0,0).$ Then \[B = (-a, -a), F = (0,c), BF: y = x (1 + \frac {c}{a})+c.\] \[E=(b, -a),  AE: y = -\frac {a}{b}x.\] \[D = (-a,0), C= (b,c), CD: y = c \frac {x+a}{a+b}.\] \[X = CD \cap AE \cap BF = (-bk, ak),\] where $k=  \frac {c} {a+b +{\frac {bc}{a}}}$ as desired.

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