Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel. The lines $AB$ and $CD$ intersect at point $P$. The lines $AD$ and $BC$ intersect in point $Q$. The bisector of the angle $\angle DPA$ cuts the line segment $BC$ and $DA$ in the points $E$ and $G$, respectively. The bisector of the angle $\angle AQB$ cuts the line segments $AB$ and $CD$ in the points $H$ and $F$.

Now it seems as if the quadrilateral $EFGH$ is a always a rhombus. I intend to prove this.

This is a sketch of the problem

Maybe anyone has a checklist or any idea to begin with.

  • 2
    $\begingroup$ @Mathematic.al So is the $ ABCD$ quadrilateral given as cyclic? $\endgroup$ – Narasimham Nov 17 '18 at 21:03
  • $\begingroup$ Yes, this is why I've constructed the circle $\endgroup$ – calculatormathematical Nov 17 '18 at 23:53

We first show that $EG \perp HF$. Let $E'$ and $G'$ be the intersections of line $GE$ with the circle so that $G$ is between $G'$ and $E$. Similarly, define $H'$ and $F'$. For any two points $X,Y$ on the circle, let $XY$ denote the radian measure of the shorter arc connecting them (sorry, I couldn't figure out the arc command here). By the assumptions, we have $$\begin{align}\newcommand{arc}[1]{\overset{\mmlToken{mo}{⏜}}{#1}}\arc{H'E'}+\arc{G'F'}&=\arc{H'B}+\arc{BE'}+\arc{G'D}+\arc{DF'}=(\arc{H'B}+\arc{DF'})+(\arc{BE'}+\arc{G'D})\\&=(\arc{CF'}+\arc{AH'})+(\arc{E'C}+\arc{AG'})=\arc{E'C}+\arc{CF'}+\arc{H'A}+\arc{AG'}\\&=\arc{E'F'}+\arc{H'G'},\end{align}$$ which implies our claim.

Let $S$ be the intersection of $EG$ and $HF$. Now, in triangle $\triangle PHF$ the angle bisector of $\angle P$ is perpendicular ot $HF$, hence it is an isosceles triangle, hence $S$ is the midpoint of the side $HF$. Similarly, $S$ is the midpoint of side $EG$. So the quadrilateral $EFGH$ has perpendicular diagonals that bisect each other. This happens only if $EFGH$ is a rhombus.

  • $\begingroup$ I have added the arc command to your answer. I hope that you don't mind and that the edit is to your liking. $\endgroup$ – Batominovski Nov 17 '18 at 22:05
  • $\begingroup$ @Batominovski looks good thank you. $\endgroup$ – Marco Nov 18 '18 at 2:49

Here is an alternative way to show that $EG\perp HF$. In fact, I shall verify that, for any convex quadrilateral $ABCD$, the quadrilateral $EFGH$ is a rhombus if and only if the quadrilateral $ABCD$ is cyclic. Without loss of generality, suppose that the configuration of points $P$ and $Q$ are as in the OP's figure (that is, $P$ and the segment $AD$ are on the opposite side of the line $BC$, and $Q$ and the segment $CD$ are on the opposite side of the line $AB$).

Let $EG$ and $FH$ meet at $S$. Write $\alpha$, $\beta$, $\gamma$, and $\delta$ for the angles $\angle DAB$, $\angle ABC$, $\angle BCD$, and $\angle CDA$, respectively. Then, $$\angle CQD=\pi-\gamma-\delta\text{ so }\angle SQB=\frac{\angle CQD}{2}=\frac{\pi}{2}-\frac{\gamma}{2}-\frac{\delta}{2}\,.$$ Similarly, $$\angle AQD=\pi-\alpha-\delta\text{ so }\angle SPB=\frac{\angle CQD}{2}=\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\delta}{2}\,.$$ The sum of internal angles of the quadrilateral $PSQB$ is $2\pi$, whence $$\begin{align}\angle PSQ&=2\pi-\big(\angle SQB+\angle SPB+(2\pi-\angle PBQ)\big)\\&=\angle PBQ - \angle SQB-\angle SPB\,.\end{align}$$ However, $\angle PBQ=\angle ABC=\beta$, so $$\angle PSQ=\beta-\left(\frac{\pi}{2}-\frac{\gamma}{2}-\frac{\delta}{2}\right)-\left(\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\delta}{2}\right)\,.$$ That is, $$\angle PSQ=(\beta+\delta)+\frac{\alpha+\gamma}{2}-\pi=\frac{\beta+\delta}{2}+\frac{\alpha+\beta+\gamma+\delta}{2}-\pi\,.$$ Since $\alpha+\beta+\gamma+\delta=2\pi$, we have $$\angle PSQ=\frac{\beta+\delta}{2}\,.$$

If $EGHF$ is a rhombus, then $\angle PSQ=\dfrac{\pi}{2}$, making $\beta+\delta=\pi$. Consequently, the quadrilateral $ABCD$ is cyclic. Conversely, if the quadrilateral $ABCD$ is cyclic, then $\beta+\delta=\pi$ implies that $\angle PSQ=\dfrac{\pi}{2}$, so $EG\perp HF$. The rest goes as Marco's answer.


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