*The minimum inscribed triangle when one vertex is fixed.*

We want to find the minimum inscribed triangle of the fixed large black triangle that passes through the fixed point $F$.
At the moment the red points $D$ and $E$ are arbitrary: they constitute a *positio falsa* which we want to analyse and then improve.

The perimeter of the inscribed triangle is given by the sum of the lengths of the red, orange and yellow lines. The numerical value of the perimeter is given in the top right of the diagram. Move the red points to see how the perimeter changes and the inner triangle changes.

The point $F'$ is the reflection of the point $F$ in the line $AB$ and the point $F''$ is the reflection of the point $F$ in the line $BC$. Therefore the length of $DF'$ is equal to the length of $DF$ and the length of $EF''$ is equal to the length of $EF$. Note that $F'$ and $F''$ are fixed with $F$.

Therefore the length of the perimeter is equal to the length of the path $F'DEF''$. But the shortest distance between two points is a straight line. So if we want the perimeter to be as short as possible then we should choose $D$ and $E$ so they lie on the line $F'F''$. By moving the red points confirm that the perimeter is minimised when $D$ and $E$ both lie on $F'F''$.