6
$\begingroup$

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 $DA$.

enter image description here

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.