# Line passing through vertex of triangle and intersection of diagonals trapezoid formed by base and a parallel line divides both into two equal parts.

Tl;Dr : Prove that line passing through a vertex of triangle and intersection of diagonals of trapezoid formed by base and a parallel line divides both segments into two equal parts.

I was reading about Jewish(or Coffin) problems, (Link: https://arxiv.org/abs/1110.1556). I am not able to understand the solution of problem 16. I am putting the problem and the solution here.

Problem:

You are given two parallel segments. Using a straightedge, divide one
of them into six equal parts.


Solution:

Given a segment and a parallel line we can always divide the segment
into two equal parts. Take two points on the parallel line. Together
with the ends of the segment the points form a trapezoid. Continue the
sides to form a triangle. The segment passing through the third vertex
of the triangle and the intersection of the diagonals of the trapezoid
divides both parallel segments into two equal parts.

Using the division method above, divide one of the segments into eight
equal parts. Pick six of these parts consecutively. Then perform a
homothety mapping their union onto the other segment. The center of the
homothety is the intersection of the other sides of the trapezoid
formed by the six segments and the target segment.


The part I am unable to understand is:

The segment passing through the third vertex of the triangle and the
intersection of the diagonals of the trapezoid divides both parallel
segments into two equal parts.


I tried to prove this but am unable to. I don't even know where to begin. Please help me in proving this.

## 1 Answer

Trapezoid $ABCD$, whose sides $AC$ and $BD$ extended intersect at $E$, and diagonals intersect at $F$. Prove that the line drawn from $E$ through $F$, cutting $CD$ at $H$ and $AB$ at $G$, makes $AG=GB$ and $CH=HD$. Since they share base $CD$ and lie between parallels, $$\triangle ACD=\triangle BCD$$Adding triangle CDE to both$$\triangle ADE=\triangle BCE$$Subtracting quadrilateral $ECFD$ from both$$\triangle AFC=\triangle BFD$$and since they have equal angles at $F$,$$\frac{CF}{FB}=\frac{DF}{FA}$$(If equal triangles have one angle equal, the sides about the equal angles are reciprocally proportional.) Or compounding$$\frac{CB}{FB}=\frac{DA}{FA}$$ Further$$\frac{CF}{FB}=\frac{\triangle ECF}{\triangle EFB}$$or compounding$$\frac{CB}{FB}=\frac{\triangle ECB}{\triangle EFB}$$Similarly$$\frac{DF}{FA}=\frac{\triangle EDF}{\triangle EFA}$$or compounding$$\frac{DA}{FA}=\frac{\triangle EDA}{\triangle EFA}$$Therefore$$\frac{\triangle ECB}{\triangle EFB}=\frac{\triangle EDA}{\triangle EFA}$$And since $\triangle ECB=\triangle EDA$, therefore$$\triangle EFB=\triangle EFA$$And since they are equal and on common base $EF$, triangles $EFB$ and $EFA$ must lie between parallels through $A$ and $B$ that are equidistant from $EF$. But triangles $AFG$ and $BFG$ lie between those same parallels and share base $FG$. Therefore triangles $FGA$ and $FGB$ are equal. Adding these to equal triangles $EFA$ and $EFB$, the whole triangles $EGA$ and $EGB$ are equal. And since they have a common altitude, their bases are equal$$AG=GB$$And by similar triangles$$CH=HD$$

• by saying that two triangles are equal, do you mean to say that they are congruent? – Gaurav Kumar Sep 14 '17 at 15:37
• No, just that they have the same area. – Edward Porcella Sep 14 '17 at 15:43