Prove: Viviani’s theorem

Please,take a look at the Picture.

It may be easiest to first prove it for the case where the "interior" point is actually on one of the edges. Then the general case follows by cutting out a smaller equilateral triangle so that the interior point lies on the edge of it.

So assume the point $X$ is on an edge $\overline{AB}$ of equilateral $\triangle ABC$. Drop the two altitudes to get two points $Y$ on $\overline{AC}$ and $Z$ on $\overline{BC}$, and let $M$ be the midpoint of $\overline{BC}$. Then $\triangle XAY\sim \triangle XBZ \sim \triangle ABM$...

Edit: Since you only want a hint I've removed the concluding lines.

This is problem 189 in the book.

Problem 187 is

In an isosceles triangle, the sum of the distances from each point of the base to the lateral sides is constant, namely it is congruent to the altitude dropped to a lateral side.

We will apply problem 187 (without proof).

From $D$, draw the line parallel to $AB$. Consider the truncated equilateral triangle. Hence, by problem 187, $DF + DG = AI$.
Hence, $ DE + DF + DH = DE + AI = IH + AI = AH$.