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I was searching things about the diagonal triangle of a quadrilateral. I found in MathWorld's "Diagonal Triangle" entry the following question:

If the quadrangle is a cyclic quadrilateral, then the circle is the polar circle of the diagonal triangle.

How can I prove that? I tried using the fact that is cyclic, but it doesn't leave me to an answer.

Blue
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Nana
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  • You should recall what the polar circle of a triangle is: https://en.wikipedia.org/wiki/Polar_circle_(geometry) and https://mathworld.wolfram.com/PolarCircle.html – Jean Marie Dec 05 '20 at 14:51

1 Answers1

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Here is a possible solution. First, let us insert a picture with labelled vertices:

Mathematics stackexchange 39331113

We use it for several partial results, some of them involve only a few vertices, please draw the simplified picture on paper.

Proposition:

Let $ABCD$ be a cyclic quadrilateral. Let $O$ be the center of the circle $(ABCD)$. Let $Q$ be the intersection $AD\cap BC$. Assume it is an exterior point to the circle $(O)=(ABCD)$. We denote the inversion centered in $O$ that invariates pointwise the circle $(O)$ by a star. So $Q^*$ is the inverse of $Q$, it can be constructed by drawing the tangents from $Q$ to $(O)$, let $Q_1,Q_2$ be the tangent points, then $Q^*=OQ\cap Q_1Q_2$.

Let $P,R$ be the intersections $P=AC\cap BD$ and $R= AB\cap CD$.

Also draw $X=RP\cap QAD$, $Y=RP\cap QBC$.

Then we have:

  • (a) The line $Q_1Q_2$ is the polar of $Q$.
  • (b) $Q$ is the pole of $Q_1Q_1$.
  • (c) $X,Q$ are harmonic conjugates w.r.t. $A,D$.
  • (d) $Y,Q$ are harmonic conjugates w.r.t. $B,C$.
  • (e) $XY$ is the polar of $Q$.
  • (f) $Q_1,Q_2;Q^*;X,Y;P,R$ are colinear.

Proof: Some parts are simple, they are enumerated to have the steps of the result displayed clearly.

(a), (b) are standard.

(c) follows by applying the theorems of Ceva and Menelaus in $\Delta RAD$: $$ \begin{aligned} -1 &= \frac{XA}{XD}\cdot \frac{CD}{CR}\cdot \frac{BR}{BA}\ , &&\text{ Ceva w.r.t. the point $P$, and cevians $RPX$, $APC$, $DPB$,} \\ +1 &= \frac{QA}{QD}\cdot \frac{CD}{CR}\cdot \frac{BR}{BA}\ , &&\text{ Menelaus w.r.t. the secant line $QBC$.} \end{aligned} $$ (d) is (c) for the situation obtained by exchanging $A\leftrightarrow B$, $C\leftrightarrow D$.

(e) From (c), (d) both $X,Y$ are on the polar of $Q$, so this polar line is $XY=XYPR=PR$.

$\square$

Proposition: $QR$ is the polar of $P$.

Proof: $P$ lies on the polar of $Q$, so $Q$ lies on the polar of $P$, La Hire's theorem. Similarly, $R$ lies on the polar of $P$. So $QR$ is the polar of $P$.

$\square$

Note: The parts (a), (b) of the first proposition are not needed, but this is the way i constructed $Q^*$ in geogebra.

dan_fulea
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