For a point K inside a triangle show an equality

Now from your last equation, call

$M=ah_a=bh_b=ch_c$.

Then you have that $\frac{ad_a+bd_b+cd_c}{M}=1$, that is, $\frac{ad_a}{M}+\frac{bd_b}{M}+\frac{cd_c}{M}=1$, that is, $\frac{ad_a}{ah_a}+\frac{bd_b}{bh_b}+\frac{cd_c}{ch_c}=1$, which is what you want.

Observe that, $\frac {d_{a}}{h_{a}}=\frac{\frac12\times a\times d_{a}}{\frac12\times a\times h_{a}}=\frac{Area(KBC)}{Area(ABC)}$.

Similarly, $\frac{d_{b}}{h_{b}}=\frac{Area(KCA)}{Area(ABC)}$ and $\frac{d_{c}}{h_{c}}=\frac{Area(KAB)}{Area(ABC)}$.

Adding these three ratios will give the desired result.