Don't understand the solution of this geometry problem

Question

In this question, AF cuts the parallelogram ABCD. BE, CF, DG are $\perp$s from the other vertices to AF. LM is also a $\perp$ where L is the intersection of 2 diagonals in the parallelogram. Prove CF = BE - DG.

Now the solution is as follow:

In the trapezoid BEDG, L is the mid-point of BD and LM is $\parallel$ BE and DG. So M is the mid-point of the diagnoal GE.

• I don't understand this part.

Since M is the mid-point of the diagonal, LM = $\frac{1}{2}(BE-DG)$.

• I also don't understand this part.

I find this difficult to understand in a 3D context. Can someone understand the solution and explain it in another way to me?

Answer 1

Let $GDPE$ be a parallelogram(rectangle) and $LM\cap DP=\{Q\}$.

Thus, since $L$ is a mid-point of $DB$, we obtain: $$LM=LQ-QM=\frac{1}{2}PB-DG=\frac{1}{2}(BE+PE)-DG=\frac{1}{2}(BE-DG).$$ Now, use that $CF=2LM.$

Answer 2

An easy way to understand it. Consider all possible segments that connects the lines $AF$ and $BD$ and that are perpendicular to $AF$. These segments are clearly parallel among them, and the segments $BE$ and $DG$ are two of these. Now focus on the behaviour of these segments moving from $BE$ to $DG$.

Starting from $BE$ and moving towards the left in your figure, the length of these segments decreases, reduces to a single point (the crossing point between $AF$ and $BD$) and then increases, until we arrive to $DG$. This change of dimension occurs progressively in a linear manner. So, the length of the "central" segment (which in this case is $LM$) is necessarily the average between the length of the initial and final segment. To show that $LM$ is the central segment, note that $L$ is the midpoint of $DB$, because it is the intersection point of the diagonals $DB$ and $AC$ of the parallelogram $ABCD$ (this is a well known property of parallelograms: the diagonals intersect each other at the half-way point). As a result, since $DL=LB$, we also have $GM=ME$, as $GM$ and $ME$ are the projections of $DL$ and $LB$ on $AF$. Then, $M$ is the midpoint of $GE$.

In this case, since the length of our segments decreases to zero and then increases, we have to consider the length of the final segment $BD$ as negative (this can be easily understood hypothesizing the case in which the final segment is equal to the initial one: their average would be zero and accordingly the central segment would be reduced to a point). So we have $$LM=(BE-DG)/2$$