Prove that in a quadrilateral, the lines joining the midpoints of the opposite sides and the midpoints of the diagonals are concurrent.

We construct an arbitrary quadrilateral $ABCD$ with $E, F, G$ as the midpoints of $AB, BC, CD$. Let $H, I$ be the midpoints of $AC, BD$. Let $EG, HI$ intersect at $J$. Let the line joining $F, J$ meet $AD$ at $K$. We will prove that $K$ is the midpoint of $DA$.

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Joining $KG, GF, FE, EK$, it quickly becomes clear that the above is only true if $KGFE$ is a parallelogram, which in turn, is only true if $EJ = JG, KJ = JF$. Proving the first equality is easy.

In $\Delta ABC, EH || BC, 2\cdot EH = BC$. Likewise, in $\Delta DBC, IG || BC, 2\cdot IG = BC$. Therefore, $IG||EH, IG=EH$. Therefore, $EHGI$ is a parallelogram and $EJ = JG$. Even after numerous efforts I wasn't able to prove the second equality.

I noticed that this was because I was not utilizing the fact that $F$ is the midpoint of $BC$ and that $FK$ is the straight line.

So, to utilize those facts, I considered $\Delta HKJ, \Delta JFI$. Proving these are congruent will prove our conjecture. Now, we can use the fact that $FK$ is a straight line by saying, that $\angle HJK = \angle IJF$. Also, since $EHGI$ is a parallelogram, $HJ = JI$. Now we need only one more equivalence to prove congruency. I wasn't able to find this.

A way to utilize the fact that $F$ is the midpoint of $BC$ is by noticing that $EHCF, IGFC, AHFE, FIDG$ are all parallelograms. I have, however, no idea how to use these in the proof.

I think I'm forgetting something. Because in each approach I take, there is always a single piece that is missing. If anybody could point out what this 'piece' is, I would be grateful. I would appreciate solutions that are related to the approaches described above.

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    $\begingroup$ The result certainly seems true when you squint your eyes and imagine that $ABCD$ is in fact a tetrahedron. If you already knew that lines joining midpoints of opposite edges of a tetrahedron concur, you'd be done ... but I don't think this is a well-known fact (although it's about as easy to show as the fact that the midpoint polygon of a quadrilateral is a parallelogram). $\endgroup$ – Blue Dec 26 '13 at 11:09
  • $\begingroup$ Don't you want to use vector? Using vector is better, I think. $\endgroup$ – mathlove Dec 26 '13 at 11:22
  • $\begingroup$ The apparent parallelograms ($KGFE$, $KHFI$, $HEIG$) are actual parallelograms, because they're the midpoint polygons of various quadrilaterals (which may be "bow-tie" quads, but that's okay). $\endgroup$ – Blue Dec 26 '13 at 11:22
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    $\begingroup$ If will probably easier if you set $K$ as the midpoint of $AD$ and show all three lines $EG$, $FK$ and $HI$ intersect at the same point $J$. When $K$ is the mid-point of $AD$, $EFGH$ is a parallelogram. So $EG$ and $FK$ intersect at a point $J$ which is the midpoint of $EG$ and $FK$. Similarly, $FIKH$ is another parallelogram, so $FK$ and $HI$ intersect at another point $J'$ which is the midpoint of $FK$ and $HI$. Since both $J$ and $J'$ are midpoint of $FK$, $J = J'$ and we are done. $\endgroup$ – achille hui Dec 26 '13 at 11:23
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    $\begingroup$ @Gerard, it is usually more complicated. But in this case, if you think in terms of vector like others suggested, then it is sort of obvious $\vec{J} = \frac14 \left(\vec{A}+\vec{B}+\vec{C}+\vec{D}\right)$, so you know the complicated way will give you an answer. You just look for geometric constructs to help you translate a vector based proof to a pure geometric proof. Guessing the right answer usually solves half of the problem. $\endgroup$ – achille hui Dec 26 '13 at 11:37

I'm going to write an answer using vector.

Let $O$ be the intersection point of $AC, BD$.

Let $$\vec{OA}=\vec{a}, \vec{OB}=\vec{b}, \vec{OC}=k\vec{a}, \vec{OD}=l\vec{b}$$ where $k,l\lt 0.$

Letting $E,F,G,H$ be the midpoint of $AB, BC, CD, DA$ respectively, we have $$\vec{OE}=\frac{1}{2}\vec a+\frac 12\vec b,\vec{OF}=\frac 12\vec b+\frac k2\vec a, \vec{OG}=\frac k2\vec a+\frac l2\vec b, \vec{OH}=\frac 12\vec a+\frac l2\vec b.$$

Letting $I$ be the intersection point of $EG, FH$, there exist $m,n$ such that $$\vec{EI}=m\vec{EG}, \vec{FI}=n\vec{FH}.$$

The former gives us $$\vec{OI}-\vec{OE}=m\left(\vec{OG}-\vec{OE}\right)\iff \vec{OI}=(1-m)\vec{OE}+m\vec{OG}=\frac{1-m+mk}{2}\vec a+\frac{1-m+ml}{2}\vec b.$$

The latter gives us $$\vec{OI}-\vec{OF}=n\left(\vec{OH}-\vec{OF}\right)\iff \vec{OI}=(1-n)\vec{OF}+n\vec{OH}=\frac{k-kn+n}{2}\vec a+\frac{1-n+nl}{2}\vec b.$$

Now since $\vec a$ and $\vec b$ are linearly independent, the following has to be satisfied :

$$\frac{1-m+mk}{2}=\frac{k-kn+n}{2}\ \text{and} \frac{1-m+ml}{2}=\frac{1-n+nl}{2}.$$ These give us $m=n=1/2$ since $(k,l)\not=(-1,-1).$ Hence, we get $$\vec{OI}=\frac{k+1}{4}\vec a+\frac{l+1}{4}\vec b.$$

On the other hand, letting $P,Q$ be the midpoint of $AC, BD$, we have $$\vec{OP}=\frac{k+1}{2}\vec a, \vec{OQ}=\frac{l+1}{2}\vec b.$$

Finally, we can lead $$\vec{PI}=\frac 12\vec{PQ}.$$ Since this represents that $I$ is on the line $PQ$, we now know that we get what we want. Q.E.D.

P.S. If $(k,l)=(-1,-1)$, then $ABCD$ is a parallelogram, which is an easy case.

  • $\begingroup$ So, $\vec{OC}$ is just the scaled version of $\vec{OA}$? That doesn't make sense. I mean, the two clearly have a different direction, don' they? $\endgroup$ – Gerard Dec 26 '13 at 14:48
  • $\begingroup$ In my answer, $O$ is the intersection point of $AC, BD$. $\endgroup$ – mathlove Dec 26 '13 at 14:49
  • $\begingroup$ Oh, you should mention that. I thought $O$ referred to the origin. $\endgroup$ – Gerard Dec 26 '13 at 14:52
  • $\begingroup$ I think so. sorry. I added. $\endgroup$ – mathlove Dec 26 '13 at 14:53
  • $\begingroup$ How did you get $\vec{OI}=\frac{k+1}{4}\vec a+\frac{l+1}{4}\vec b$? Forgive me, but I'm new to vector geometry. $\endgroup$ – Gerard Dec 26 '13 at 15:55

Using the same diagram as above in the question, I will go with this way:

FHKI is a parallelogram [HK || CD by applying equal intercepts theorem in the triangle ACD. Similarly FI || CD; and IK || AB, FH || AB. Since opposite pairs of sides are parallel, FHKI is indeed a parallelogram ]

Similarly EIGH is parallelogram.

Now the key argument:

Let the diagonals HI and EG of parallelogram EHGI meet at point J. Then J bisects HI.

Similarly, let the diagonals HI and FK of parallelogram FHKI meet at point J'. Then J' bisects HI.

But there can be only one midpoint of a line [which is HI in current situation], hence J = J'.

Hence EG, FK, HI are concurrent.


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