# Prove: Viviani’s theorem

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.

• There is a proof via similarity of triangles. Would that be in line with what you want? (I'm not sure where/how midline theorems were introduced) – Calvin Lin Mar 28 at 20:28
• I have his book (not sure about edition number). Do you have a page reference or section name so that I can take a look? – Calvin Lin Mar 28 at 20:30
• Section 13, Problem 189. – Hassan Ashraf Mar 28 at 20:38
• Page 76, and thanks for your help. – Hassan Ashraf Mar 28 at 20:38
• I believe that you were meant to use problem 187, as that lends nicely to 189 (which is this problem). – Calvin Lin Mar 28 at 23:05

Please,take a look at the Picture.

• This brilliant, thank you! But Is KE = h, by construction or is it proved? – Hassan Ashraf Mar 28 at 21:19
• Well , it is easily proved , since E is symmetrical to A relative to axis CB. Therefore, you may consider it proved or proved by construction, whatever. – ole Mar 28 at 21:25

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 a cool proof indeed, but sadly I don't think it was the intentional one, since, Similarity and Proportion is two chapters ahead of this problem. I have tried considering that the point X is on one altitude, we'll get the two distances are equal to a part of the altitude (Since they will form a 30-60-90 triangle), and the third distance is basically the remaining part of the altitude. But I couldn't gerenalize it using the the same reasoning. – Hassan Ashraf Mar 28 at 21:04
• Hmm, without similarity in your toolbox I don't know what to tell you. Regardless, thanks for bringing this beautiful theorem to my attention. – Jair Taylor Mar 28 at 21:12
• And thank you for providing this beautiful proof. – Hassan Ashraf Mar 28 at 21:14

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, $$DE + DF + DH = DE + AI = IH + AI = AH$$.

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