mg.metric geometry – Collinearity in tangential pentagon

I am looking for a proof of the following claim:

Given tangential pentagon. Touching point of the incircle and the side of the pentagon,the vertex opposite to that side and the intersection point of diagonals drawn from endpoints of that same side are collinear.

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The GeoGebra applet that demonstrates this claim can be found here.

mg.metric geometry – Necessary and sufficient condition for tangential polygon to be cyclic

Can you prove or disprove the following claim?

Claim. Let $A_1,A_2, ldots ,A_n$ be the vertices of an $n$-sided tangential polygon and let $B_1,B_2, ldots ,B_n$ be the contact points of the inscribed circle and polygon sides such that $B_1$ lies on $A_1A_2$, $B_2$ lies on $A_2A_3$ ,etc. Denote by $H_1,H_2, ldots,H_n$ the orthocenters of the triangles $triangle A_1B_1B_n$, $triangle A_2B_2B_1$,….,$triangle A_nB_{n}B_{n-1}$ . Then the polygon is cyclic if and only if $H_1,H_2;ldots ,H_n$ are concyclic.

Picture for the case $n=6$:

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GeoGebra applets that demonstrate this claim can be found here , here and here.

How can we calculate this tangential differential?

Let $tau>0$, $dinmathbb N$ and $T_t$ be a $C^1$-diffeomorphism on $mathbb R^d$ for $tin(0,tau)$ with $T_0=operatorname{id}_{mathbb R^d}$. Assume $$T_t(x)=x+int_0^tv(s,T_s(x)):{rm d}s;;;text{for all }(t,x)in(0,tau)tag1$$ for some sufficiently nice $v:(0,tau)timesmathbb R^dtomathbb R^d$.

Let $mathcal M$ denote the set of embedded $C^1$-submanifolds of $mathbb R^d$ with boundary and $z:{partialOmega:Omegainmathcal M}tomathbb R$. We say that $z$ has a material derivative at $partialOmega$ in direction $v$ if $${rm d}y(Omega;v)(x):=left.frac{rm d}{{rm d}t}yleft(T_t(partialOmega)right)(T_t(x))right|_{t=0}tag2$$ exists for all $xinpartialOmega$.

If $M$ is an embedded $C^1$-submanifold of $mathbb R^d$ with or without boundary, let $operatorname P_M(x)$ denote the orthogonal projection of $mathbb R^d$ onto the tangent space $T_x:M$ for $xin M$ and if $f:Mtomathbb R$ is $C^1$-differentiable at $xin M$ define $${rm D}_Mf(x):=T_x(f)circoperatorname P_M(x)tag3.$$

Now assume $$z(partialOmega)=left.fright|_{partialOmega}tag4$$ for some $fin C^1(mathbb R^d)$. Why does it follow that $${rm d}z(partialOmega;v)(x)={rm D}_{partial M}f(x)v(0,x)tag5$$ for all $xinpartial M$?

Since, by assumption, $$yleft(T_t(partialOmega)right)(T_t(x))=f(T_t(x));;;text{for all }xinpartialOmegatag6,$$ shouldn’t we get $${rm d}z(partialOmega;v)(x)={rm D}f(x)v(0,x)tag7$$ for all $xinpartial M$?

The claim $(5)$ is made in Lemma 4.3.8-ii-II in this paper.