Linear relation inside of a triangle Let $d(P,l)$ be distance between point $P$ and line $l$. Inside triangle $ABC$ there are points $D$ and $E$, for which $$ d(D,AC)+d(D,BC)=d(D,AB), $$ $$ d(E,AC)+d(E,BC)=d(E,AB). $$ Prove that $d(X,AC)+d(X,BC)=d(X,AB)$ for any point $X$ on segment $DE$. Looks like there is a laborious analytical proof relying on the coordinate method. But could there be an easier solution, that is also purely geometric?

The proof is as simple as it gets:

![enter image description here](

It's given that:

$$d_1+d_2=d_3$$

$$e_1+e_2=e_3$$

Introduce: $DX=u$ for some point $X\in DE$.

$$x_1=d_1+\frac{u}{DE}(e_1-d_1)$$

$$x_2=d_2+\frac{u}{DE}(e_2-d_2)$$

$$x_1+x_2=d_1+\frac{u}{DE}(e_1-d_1)+d_2+\frac{u}{DE}(e_2-d_2)=$$

$$x_1+x_2=(d_1+d_2)+\frac{u}{DE}((e_1+e2)-(d_1+d2))=$$

$$x_1+x_2=d_3+\frac{u}{DE}(e_3-d_3)=$$

$$x_1+x_2=x_3$$

Yes, it's all about "linearity".