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I am self-studying Euclidean Geometry, and I want to solve the following exercise.

Let $ABC$ be an equilateral triangle with height $h$, and $P$ is a point in its interior. If $x,y,z$ denote the distances from $P$ to its sides, then $x+y+z=h.$

I would appreciate any help with this question.

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2 Answers

up vote 3 down vote accepted

Let $a$ be the length of a side of our equilateral triangle. Join $P$ to the corners. This divides our triangle into three triangles, with areas respectively $(1/2)ax$, $(1/2)ay$, and $(1/2)az$, so combined area $(1/2)a(x+y+z)$.

But the area of our equilateral triangle is $(1/2)ah$. It follows that $$\frac{1}{2}a(x+y+z)=\frac{1}{2}ah.$$ The result now follows by cancellation.

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If you are interested, this result is called Viviani's Theorem. –  EuYu Oct 5 '12 at 18:23
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Place the triangle in a coordinate system.

Within the triangle (as well as on its boundary), each of $x$, $y$, and $z$ is an affine function of the coordinates of each point. The sum $x+y+z$ is therefore also an affine function.

Since it is clear that $x+y+z=h$ at each of the three corners (at each corner two of them is $0$ and the third is $h$), we can find the value of $x+y+z$ anywhere in the triangle by linear interpolation between the three $h$s. But that means that $x+y+z$ is always $h$.

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