# Rhombus in a cyclic quadrilateral

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.

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

• @Mathematic.al So is the $ABCD$ quadrilateral given as cyclic? Nov 17, 2018 at 21:03
• Yes, this is why I've constructed the circle Nov 17, 2018 at 23:53

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.
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.