The task is as follows:
ABCD is a convex quadrilateral with points E, F, G and H on its sides AB, BC, CD and DA so that they divide the respective side at the ratio of the adjacent sides, so $$\frac {AE}{EB}=\frac {DA}{BC}, \frac {BF}{FC}=\frac {AB}{CD}, \frac {CG}{GD}=\frac {BC}{DA}, \frac {DH}{HA}=\frac {CD}{AB}.$$ Show that ABCD is a tangential quadrilateral if and only if EFGH is a cyclic quadrilateral.
I have already shown that if ABCD is a tangential quadrilateral, EFGH is a cyclic quadrilateral. But I have not succeeded in proving the reverse direction yet. I cannot figure out what to make out of these ratios. Thank you in advance!
With some manipulation, we obtain $$\frac{CG}{CF}=\frac{DG}{DH}=\frac{AE}{AF}=\frac{BE}{BF}=\frac{AB+CD}{AD+BC}.\tag{1}$$
Without loss of generality, suppose that $AB+CD\ge AD+BC$. Then from (1), we get that the green angles are not larger than the red angles. However, their corresponding sums are both $180^\circ$ because $EFGH$ are cyclic. Thus all the ratios in (1) are equal to $1$ and $ABCD$ is a tangential quadrilateral.