Prove that the midpoints of the sides of a quadrilateral lie on a circle if and only if the quadrilateral is orthodiagonal.

My effort: enter image description here

For the if part i did as follows:

Given orthodiagonal quadrilateral $ABCD$ we draw diagonals $AC$ and $BD$,from the points of intersection of the diagonals we then draw segment $OP$ where $P$ is the midpoint of $AD$.

Since $\Delta AOD$ is a right triangle we have that $OP$ is its median,hence $OP=AP=PD$.

Now, given that $\Delta APR \cong \Delta OPR$, we see by simmetry that we have $AX=XO$, hence $PX$ is the altitude of isosceles triangle $APO$ ,and from this fact it follows also that $\Delta APX \cong \Delta OPX $.

By the same argument $PY$ is the altitude of isosceles $DPO$,therefore we have $PY=XO$ and $\Delta DPY \cong \Delta OPY \cong \Delta APX \cong \Delta OPX$.

Finally if we consider that $\angle XPO =\angle PDY$ and $\angle OPY =\angle DPY $ we

have that $\angle XPO +\angle OPY =\angle APX +\angle OPY= \angle RPT=90^\circ $.

In this way I can show that quadrilateral $PRST$ is cyclic by showing that also the opposite angle $RST$ is $90^\circ$.


Is this line of reasoning (for sure redundant in some point) okay in general ?In particular can anyone give me some hints on how to approach the only if part ? Also if you can provide any advice on better ways to approach the problem it would be appreciated.

Thanks in advance and forgive me for any english mystakes or if the phrasing wasn't really clear(feel free to edit in case).

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    $\begingroup$ As far as I understand, for the only if part, you are saying that since you have that "orthodiagonal" $\Rightarrow$ "the fact" $\Rightarrow$ "cyclic", you have that "not orthodiagonal" $\Rightarrow$ "not cyclic". I don't know why you can say so. $\endgroup$ – mathlove Oct 20 '15 at 17:39
  • $\begingroup$ you're right,i will work on that part more.I think i have to seek for a contradiction somewhere,could you give me any hint?Also do you think that the first part is fine? btw thanks for your time $\endgroup$ – Nameless Oct 20 '15 at 17:44

For the only if part :

Let us prove that if a quadrilateral is not orthodiagonal, then the midpoints of the quadrilateral are not concyclic.

Let me use the figure you drew. Let $\angle{AOD}=\alpha$ where $\alpha\not=90^\circ$.

Since $\angle{PYD}=\alpha$ and $\angle{AXP}=\alpha$, we have $$\angle{XPY}=360^\circ-\angle{PYO}-\angle{PXO}-\angle{XOY}=\alpha.$$

Similarly, we have $\angle{RST}=\alpha$. So, $\angle{XPY}+\angle{RST}=2\alpha\not=180^\circ$.

Hence, $PRST$ is not concyclic.

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    $\begingroup$ @Jhon: We can say $AX=XO$ from $\triangle{AXP}\cong\triangle{OXP}$, can't we? $\endgroup$ – mathlove Oct 20 '15 at 19:23
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    $\begingroup$ @Jhon: I think that we can prove $\triangle{AXP}\cong\triangle{OXP}$ without having $X$ is the midpoint : $AP=OP, XP=XP, \angle{AXP}=\angle{OXP}=90^\circ$. $\endgroup$ – mathlove Oct 20 '15 at 19:28
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    $\begingroup$ @Jhon: Do you see that we have that $PR$ is parallel to $DB$? $\endgroup$ – mathlove Oct 20 '15 at 19:32
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    $\begingroup$ @Jhon: Maybe you have a typo, but it's fine if everything is fine to you. $\endgroup$ – mathlove Oct 20 '15 at 19:37
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    $\begingroup$ @Jhon: Do you mean that $PA\color{red}{R}$ and $DAB$ are similar? $\endgroup$ – mathlove Oct 20 '15 at 19:43

Given: ABCD is a quadrilateral.

First, note that by joining the midpoints of sides of a quadrilateral will always result in a parrallelogram.

So, in the given diagram, we get to know PRST is atleast a parallelogram.

Now, it is given diagonals are orthogonal, i.e. they interesect at 90 degrees.

PX is parallel to YO,and XR is parallel to OB.

So the diagonals of quadrilateral ABCD divides the parallelogram into four quadrilateral in which all their individual interior angles is equal to 90 degrees

So angle P= angle R =angle S =angle T=90 degrees

So, PRST is atleast a rectangle since each angle is 90 degrees. We know all rectangles are concylic,since opposite angles are supplementary.



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