Given a triangle construct a point satsfying the mentioned properties.

Question

Given any triangle $ABC$ with $BC$ midpoint being $D$ and also given that $E$ is the midpoint of of $AB$ , construct a point $X$ such that $XB = BD$ along the side BC (not along the $\vec{BC}$ direction but other direction ) $X \neq D$ using only straightedge. (Give the motivation behind your construction too )

My try : For the point $X$ we most likely need to make/find two lines which intersect at that required point . One line can be the side $BC$ the other line must be not parallel to it . But now i am having trouble finding that line .

Answer 1

Background

Straightedge-only constructions are known as Steiner constructions. By the Poncelet–Steiner theorem, any compass-and-straightedge construction may be performed with a straightedge only, so long as a circle and its centre are given. The term “Steiner construction” seems to be used for any straightedge-only constructions, though, whether or not they make use of this circle.

In this case, we need just one construction to solve the problem. Okay, we need three, but two of them are standard Euclidean constructions that don’t need a compass, namely “construct a line through two points” and “construct the point that is the intersection of two lines”.

The third construction is “construct a parallel to a given line (segment) through a given point”. This doesn’t require a circle, but does require the midpoint of the line segment. I expect that that’s why point $E$ is given in the problem.

Construction

I found this method on Wikipedia, Poncelet–Steiner theorem § Parallel of a line having a collinear bisected segment. Here I give it slightly more briefly.

Given a line segment $\overline{AB}$ with its midpoint $M$, and a point $P$ not collinear with $AB$, construct $Q$ such that $AB \parallel PQ$:

1. Construct $AP$ and $BP$.
2. Let $R$ be an arbitrary point on $AP$.
3. Construct $BR$ and $MR$.
4. Mark $X$ at the intersection of $MR$ and $BP$.
5. Construct $AX$.
6. Mark the required point $Q$ at the intersection of $AX$ and $BR$.

Solution

The given points $A$, $B$, $C$, $D$, and $E$, the midpoint of $AC$ (call it $F$), and the required point $X$, are grid points of a parallelogram grid. The grid lines are parallel to $AB$ and $BC$, and the cells are congruent to $BDFE$. We only need to construct two of these grid lines to locate $X$.

Given $\triangle ABC$, with midpoints $D$ on $\overline{BC}$ and $E$ on $\overline{AB}$, construct $X$ such that $XBDC$ are collinear and equally spaced:

1. Construct line 1 parallel to $BC$ through $A$.
2. Construct line 2 parallel to $AB$ through $D$.
3. Mark $Y$ at the intersection of lines 1 and 2.
4. Construct line $EY$.
5. Mark the required point $X$ at the intersection of $EY$ and (extended) $BC$.