Contents
Euclidean Geometric Approach
Let's assume we have some arbitrary triangle and we would like to determine what is the shortest distance from a point to the opposite side (basically the height of that point above the opposite side) we can view it for example as follows:
a
b
h
c1
c2
c
Where a,b and c are all known values we would like to determine what the value of h is. We already know that c1+c2=c (h cuts c into c1 and c2).
$$
\begin{aligned}
c & = c1 + c2\\
h^2 & = a^2 - c1^2\\
h^2 & = b^2 - c2^2\\
a^2 - c1^2 & = b^2 - c2^2\\
a^2 - b^2 & = c1^2 - c2^2\\
& = (c1+c2) \cdot (c1-c2)\\
& = c \cdot (c1-c2)\\
c1 - c2 & = {{a^2 - b^2} \over c}\\
c1 - (c - c1) & = {{a^2 - b^2} \over c}\\
2 \cdot c1 & = {{a^2 - b^2 + c^2} \over c}\\
c1 & = {{a^2 - b^2 + c^2} \over {2 \cdot c}}\\
h^2 & = a^2 - c1^2\\
& = a^2 - \Big({{a^2 - b^2 + c^2} \over {2 \cdot c}}\Big)^2\\
\end{aligned}
$$
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Algebraic Geometric Approach
Let's put the triangle on a cartesian plan and determine the height based on the positions of the corners on the plane.
a
b
h
c1
c2
c
(0,0)
(c,0)
(c1,h)
We know that a circle with center at point (0,0) and radius a and a circle with center at (c,0) and radius b will intersect at (c1,h), thus:
$$
\begin{aligned}
a^2 & = c1^2 + h^2 \\
b^2 & = (c1-c)^2 + h^2 \\
& = c1^2 - 2 \cdot c1 \cdot c + c^2 + h^2 \\
b^2 - a^2 & = - 2 \cdot c1 \cdot c + c^2\\
c1 & = {{a^2 - b^2 + c^2} \over {2 \cdot c}}\\
h^2 & = a^2 - c1^2\\
& = a^2 - \Big({{a^2 - b^2 + c^2} \over {2 \cdot c}}\Big)^2\\
\end{aligned}
$$
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See Also:
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