$\triangle ABC$ is acute. Let $D$ lie on $AB$ such that $AC^2-BC^2=BD^2-AD^2$. Show that $CD$ is an altitude.
If we suppose $CD$ is not perpendicular to $AB$, then we have $CD_1\perp AB, D_1\in AB.$ By Pythagorean theorem we can get $AC^2=AD_1^2+CD_1^2$ and $BC^2=BD_1^2+CD_1^2$. Therefore, $AC^2-BC^2=AD_1^2-BD_1^2=BD^2-AD^2$. How to approach the problem further?
Indeed, the problem statement should be $AC^2-BC^2=AD^2-BD^2$. See the comments for more thoughts about this.
On
Apply the cosine rule to the triangles ADC and BDC
$$AC^2= AD^2 + DC^2-2AD\cdot DC\cos\angle ADC\tag1$$ $$BC^2= BD^2 + DC^2-2BD\cdot DC\cos\angle BDC\tag2$$
Note $\cos\angle ADC = -\cos\angle BDC$. Take (2)-(1) and assume the given $AC^2-BC^2=AD^2- BD^2 $to arrive at
$$AB\cdot CD \cos\angle ADC =0$$
which leads to $\angle ADC = 90$, hence $CD$ being the altitude.
It should be $$BC^2-AC^2=BD^2-AD^2.$$
Now, by the Pythagorean theorem we obtain: $$BC^2-AC^2=BD_1^2-AD_1^2,$$ which gives $$BD^2-AD^2=BD_1^2-AD_1^2$$ or $$(BD+AD)(BD-AD)=(BD_1+AD_1)(BD_1-AD_1)$$ or $$AB(BD-AD)=AB(BD_1-AD_1)$$
$$BD-AD=BD_1-AD_1$$ or $$BD-BD_1=AD-AD_1,$$ which gives $$D\equiv D_1.$$
Indeed, let $D_1$ is placed between $D$ and $B$ (the case $D_1$ is placed between $A$ and $D$ is the same).
Thus, $D$ is placed between $A$ and $D_1$ and $$BD-BD_1=AD-AD_1$$ gives $$DD_1=-DD_1,$$ which gives $$DD_1=0.$$