martes, 1 de junio de 2021

Using the half-angle formula for cosine to derive Zelich's lemma on mixtilinear incircles

Lemma (Ivan Zelich). Let $w$ be the $A$-mixtilinear incircle in $\triangle{ABC}$ touching side $AB$ at $E$, side $AC$ at $F$. Then


where $s$ is the semiperimeter.

Proof 1. The radius of $w$ inscribed in $\angle{CAB}=\alpha$ is given by


where $r$ is the inradius of the reference triangle and $\rho_a$ is the radius of $w$ (Durell and Robson 1935).

A proof of $(2)$ can be found here (see pp. 13).

The half-angle formula for cosine states that 


See here for a proof of $(3)$.

Call $I$ the Incenter of $\triangle{ABC}$ and $D$ the touchpoint between the incircle and $AC$. Denote $K$ the center of $w$. Notice that $\triangle{AID}\sim\triangle{AKF}$. Now, by similarity of triangles we have


Combining $(2)$ and $(3)$ in $(4)$ we get $(1)$.

Proof 2. We can also derive $(1)$ using the relationships $\Delta=rs$ and $\Delta=\frac{bc\sin{\alpha}}{2}$. Indeed, since 


Substituting in $(2)$ and simplifying we get 


from which the result holds. 


A proof using inversion can be found here

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