Is there a $\triangle ABC$, with $D$ on $BC$, such that $AB$, $BC$, $CA$, $AD$, $BD$, $CD$ are all integers, but only one is even? This is just a question I came up with while working through another problem.

If we have a triangle $ABC$ and we draw a cevian from $A$ to $D$ on side $BC$, is it possible to assign integer values to lengths $AB$, $BC$, $AC$, $AD$, $BD$ or $CD$ such that only one of the listed lengths has an even valued length? If it is possible, what are these lengths?

I have done some parts of this problem on my own. I figured out that this even length could only be $BC$, $CD$, or $BD$, with $BD$ and $CD$ being the same case.
With $BC$ being the only even length I computed for $BC$ using Stewart's Theorem and got
$$BC = \frac{(AC^2)(BD)+(AB^2)(CD)}{(BD)(CD)+(AD)^2}$$ where every length but $BC$ is odd. The equation seems to work but I don't know how to obtain its integer solutions.
Similarly with $BD$/$CD$ case, I applied Stewart's theorem and got $$BD=\frac{(AB^2)(CD)-(BC)(AD^2)}{(CD)(BC)-(AC)^2}$$
Once again this seems to work, but I don't know how to obtain its integer solutions.
Pretty much all I need help is verifying whether or not my work is right, and if it  is how to find the integer solutions of my provided equations. If there is a better way or my work has flaws please let me know.
 A: A simple solution found by brute force (by setting $AD=AB$):

and three more interesting ones found by exhaustive search:



A: Too long for a comment but here is food for thought.
It would appear, incorrectly, that we can find what you need if we join two dissimilar Pythagorean triples on a matching side-A. This side and be any odd number greater than one. We begin with Euclid' formula here shown as.
$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$
If we solve the A-function for k, we can find triples that match a length "A" by testing a finite range of m-values to see which ones yield integers.
\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\qquad\text{for}\qquad  \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=15\implies \lfloor\sqrt{15+1}\rfloor=4\le m \le \frac{15+1}{2} =8\quad\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$
If these two are joined at side-A, we have an acute triangle with sides $$(17,113,120)$$ with a cevian $(15)$ going from the $(17,113)$ vertex to the. 120-side, splitting it into lengths of $8$ and $112$. Unfortunately, this yields $3$ even lengths. If we were to join the even sides of two triples such as
$$(5,12,13)\qquad (35,12,37)$$ we would have $(13,37,40)$ with $40$ split into $(5,35)$. That would still leave two even lengths: the base $(40)$ and the cevian $(12)$. This probably eliminates Pythagorean triples from the solution. There are methods of finding integer cevians in a right triangle from a non-right vertex to an adjacent side but the question remains about the split-length of the target side being integers.
