By angle chasing. You can show that the incentres of ABC (K), BCD (L), B and C are cyclic because angle CLB = 180 - DCB/2 - DBC/2 = 90 + BDC/2, and similarly, angle CKB = 90 + BAC/2. But BAC = BDC. Thus CLB = CKB and so B,C,K,L are concyclic. Also if we call the incentre of ADC (M), D,C,L,M is cyclic. angle KLC = 180 - angle LBC, angle MLC = 180 - MDC, thus angle KLM = angle KBC + angle MDC = (ADC+ABC)/2 = 90, which is what we want. PS: Can someone give me a crashcourse on latex? or point to me where?

It can be proved by complex numbers... See also Paul Yiu, Notes on Euclidean Geometry, paragraph 10.2.6 (page 154); Barry Wolk, Hyacinthos message #8154, Alexey A.Zaslavsky, Hyacinthos message #8178. darij

There is the converse of this fact (from Russian Summer TST 2005): Let $ABCD$ be a quadrilateral. The incentres of the triangles $ABC, BCD, CDA, DAB$ form a rectangle. Proove, that $ABCD$ is cyclic. It's very interesting and rather difficult problem.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Le%20rectangle%20de%20Ryokan%20Maruyama.pdf Sincerely Jean-Louis

We use complex numbers. Plot the points on the unit circle. Then we can choose $a,b,c,d$ such that $A=a^2,B=b^2,C=c^2,D=d^2$, and \[I_{ABC}=-bc-ca-ab,I_{BCD}=-cd-db-bc,I_{CDA}=da+ac-cd,I_{DAB}=-ab+bd+da.\]Then we can compute \[I_{ABC}+I_{CDA}=-ab-bc-cd+da=I_{BCD}+I_{DAB}\]and \begin{align*} \left|I_{ABC}-I_{CDA}\right|^2&=\left|-bc-2ca-ab-da+cd\right|^2=\frac{1}{abcd}\left(-bc-2ca-ab-da+cd\right)\left(-ad-2bd-cd-bc+ab\right)\\ &=\frac{1}{abcd}\left(-cd-2db-bc-da+ab\right)\left(-ab-2ac-ad-bc+cd\right)=\left|-cd-2db-bc-da+ab\right|^2=\left|I_{BCD}-I_{DAB}\right|^2 \end{align*}so the incenters form a rectangle.

Labels are on the diagram. By Fact 5 $AMNB$ is cyclic. And so is $AMQD$. $\angle NMQ=\angle B/2 +\angle D/2=90^\circ$. And similarly the others.

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i only could find those right angles on the rectangle by drawing external tangents and since it is a quadrilateral it must be a square or rectangle and by length chasing interestingly we could say that that square/rectangle is definitely composed of rectangles

We use complex numbers on this one. We just easily prove this a parallelogram and then just show perpendicularity once or prove diagonals equal in length.

My solution at https://artofproblemsolving.com/community/c6t48f6h24483_abcd_cyclic_gt_incenters_of_bcd_etc_form_rectangle I hope you will enjoy it! Best regards, sunken rock

https://groups.io/g/euclid/topic/81006484#1476

Another property. https://geometry-diary.blogspot.com/2020/11/1521.html

Also China 1986 TsT