wlog let $ABCD$ be the unit square $A (0,1)$, $B (1,1)$, $C(1,0)$, $D(0,0)$ $E(a,1-a)$ with $0<a<1$ and $a\ne \frac12$ , $(BD)$ : $y=x$ orthocenter of $ABE$: $H (a,a)$ and of $ADE$: $K(1-a,1-a)$ gradient of $AH$ is $\ell_{AH} = \frac{a-1}{a-0}=\frac{a-1}{a}$ gradient of $CK$ is $\ell_{CK}= \frac{0-(1-a)}{1-(1-a)}=\frac{a-1}{a}$ finally $\ell_{AH} =\ell_{CK}$ therefore $AH \parallel CK$

Note that $K,H$ both lies on $BD$ and $\triangle ABE$ is just $\triangle ADE$ reflected over $AE$. So $K$ is just $H$ reflected over $AC$. Now we have $HK$ bisects $AC$ and $AC$ bisects $HK$ and $HK\perp BC$ so $AHCK$ is a rhombus and we're done.