Let $ABC$ be a triangle. Define $(I)$ as the incircle. Let the incircle touch $AB,BC,CA$ at $A',B',C'.$ Define $E',D',F'$ as the feet of the altitude from $A',B',C'$ on $B'C',A'C',A'B'$ respectively.Denote $I_A$ as the A excentre, $X,Y,Z$ as the touch points.
Consider triangles $A'B'C'$ and $XYZ.$ Find the length of the altitudes and simplify it.
Let $G$ be the feet of the altitude from $X$ to $YZ.$
Then $XG=XY\times \sin(XYG)=XY\times \sin(B/2).$
And $$\frac{XY}{\sin(c/2)}=\frac{s-c}{\sin(c)}\implies XG=\frac{s-c}{\sin(c)}\times\sin(c/2)\times \sin(B/2). $$
Also $XYCI_A$ is cyclic. So we can use sin rule.
We also have $EA'=A'C'\times \sin(90-C/2).$
$$\frac{A'C'}{\sin(B)}=\frac{s-c}{\sin(90-B/2)}\implies EA'=\frac{s-c}{\sin(90-B/2)}\times \sin(B)\times \sin(90-C/2)$$
And we can proceed to find others? Can someone find the other altitudes length in this way? Can we simplify them even more?
We have the following progress:
$$A'E = 2\left(s-b\right)\sin\left(\dfrac{\widehat{B}}{2}\right)\cos\left(\dfrac{\widehat{C}}{2}\right)$$
$$B'D=2(s-c)\sin(C/2)\cos(A/2)$$
$$B'D=2(s-a)\sin(A/2)\cos(B/2)$$
Define altitude from $X,Y,Z$ be $G,F,H.$ Then
$$XG=2(s-b)\sin(B/2)\cos(C/2)$$ $$ZH=2(s-c)\cos(B/2)cos(A/2)$$
We would like to suggest the method described below to determine the lengths of altitudes.
Draw the line that joins the vertex $B$ of $\triangle ABC$ to its incenter $I$. It meets the side $A'C'$ of the intouch triangle $A'B'C'$ at $V$. Since $BA' = BC'$ and $BI$ is the internal angle bisector of $\angle ABC$, we have, $$A'V = B'V\qquad\text{and}\qquad A'VB = 90^o.$$
Therefore, we shall write, $$A'V=A'B\sin\left(\widehat{VBA'}\right)=A'B\sin\left(\dfrac{\widehat{B}}{2}\right)\quad\longrightarrow\quad A'C'=2A'V=2A'B\sin\left(\dfrac{\widehat{B}}{2}\right).$$
Now, we can express $A'E\space$ as, $$A'E = A'C'\sin\left(\widehat{EC'A'}\right)= 2A'B\sin\left(\dfrac{\widehat{B}}{2}\right)\sin\left(\widehat{EC'A'}\right).$$
You have already shown that, $\measuredangle EC'A' = 90^o - \dfrac{\widehat{C}}{2}.$
Hence, we have, $$A'E = 2A'B\sin\left(\dfrac{\widehat{B}}{2}\right)\cos\left(\dfrac{\widehat{C}}{2}\right).$$
It can be shown that $A'B = s -b$, where $s = \dfrac{a+b+c}{2}.$
The expression for $A'E$ can be written as, $$A'E = 2\left(s-b\right)\sin\left(\dfrac{\widehat{B}}{2}\right)\cos\left(\dfrac{\widehat{C}}{2}\right),$$
which can be further simplified by using the property $$r= \left(s-b\right)\tan\left(\dfrac{\widehat{B}}{2}\right),$$
where $r$ is the inradius of $\triangle ABC$, to get, $$A'E=2r\cos\left(\widehat{\dfrac{B}{2}}\right)\cos\left(\widehat{\dfrac{C}{2}}\right).$$
Similarly, we can state, $$B'D=2r\cos\left(\widehat{\dfrac{C}{2}}\right)\cos\left(\widehat{\dfrac{A}{2}}\right),\quad\text{and}$$ $$C'F=2r\cos\left(\widehat{\dfrac{A}{2}}\right)\cos\left(\widehat{\dfrac{B}{2}}\right). \qquad\quad $$