Prove: Viviani’s theorem

Question

This problem is in Kiselev's Planitmetry, to prove that: In an equilateral triangle, the sum of the distances from an interior point to the sides of this triangle does not depend on the point, and is congruent to the altitude of the triangle.

After searching google for a while, I discovered that it has a name, Viviani's theorem. Anyways, the standard proof uses the concept of area, and the known formula for calculating the area of a triangle. But I don't believe that was Kiselev's intention, since, he placed the problem after the section on the midline theorems (In triangles and trapezoids), So does anybody know a way to do this? I only need a hint.

Attempt: I only found that each of these distances will be parallel to each altitude of the triangle, but couldn't use this fact in proving the theorem. In addition to that, I proved a case of the theorem, If the point lies on one of the altitudes, the proof follows from the picture.

Answer 1

Please,take a look at the Picture.

Answer 2

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.

Answer 3

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$.

Answer 4

Let x be an edges length. Sum of $AB'C$, $B'CB$ and $ABC$ triangles will be equal to triangles area which is equal to $\frac{hx}{2}$. Simplify it and boom you get $DC + EC + FC = h$ where h is height.

Solution