## 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

$$AE=AF=\frac{bc}{s}\tag{1}$$

where $s$ is the semiperimeter.

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

$$\rho_a=r\sec^2{\frac{\alpha}{2}}\tag{2}$$

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

$$\cos^2{\frac{\alpha}{2}}=\frac{s(s-a)}{bc}\tag{3}$$

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

$$\frac{r}{s-a}=\frac{\rho_a}{AF}\tag{4}$$

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

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

$$r=\frac{\Delta}{s}=\frac{\frac{bc\sin{\alpha}}{2}}{s}=\frac{bc\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}}{s}.$$

Substituting in $(2)$ and simplifying we get

$$\rho_a=r\sec^2{\frac{\alpha}{2}}=\frac{bc}{s}\tan{\frac{\alpha}{2}}=\frac{bc\rho_a}{s\cdot{AF}}$$

from which the result holds.

$\square$

A proof using inversion can be found here

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