Right Triangle Similarity

Today’s lesson in my Geometry class was on the use of the geometric mean when finding missing values of right triangles. For every right triangle, two of its altitudes are the legs and the third is perpendicular to the hypotenuse. The length of the altitude is the geometric mean of the lengths of the two segments of the hypotenuse; the length of each leg is the geometric mean of the lengths of the adjacent segment of the hypotenuse and the entire hypotenuse.

This is easier to understand using a diagram:

In the diagram, $$a^2 = h_1h_2$$ while $$l_1^2 = h_1h$$ and $$l_2^2 = h_2h$$ (such is the power of algebra and geometry that I could so quickly summarize the last sentence of the previous paragraph!).

While I find this very cool in its own right, I noticed that it’s possible to get all six of the lengths simply by knowing two of the three $$h$$ variables, since $$h = h_1 + h_2$$ (and so $$h_1 = h – h_2$$ and $$h_2 = h – h_1$$). This led me to wonder if we could figure out all six values given any two, or if there was a pair of values that wouldn’t let us do so.

Given $$l_1$$ and $$l_2$$, we can use the Pythagorean theorem: $$h^2 = l_1^2 + l_2^2$$. Since twice the area of the triangle is both $$l_1l_2$$ and $$ah$$, $$a = \frac{l_1l_2}{h}$$. Once we know that, we can calculate $$h_1$$ and $$h_2$$.

Given $$l_1$$ and $$a$$, we can use the Pythagorean theorem: $$h_1^2 = l_1^2 – a^2$$. From there, we know $$h_2 = \frac{a^2}{h_1}$$, $$h = h_1 + h_2$$, and $$l_2^2 = h^2 – l_1^2$$.

Given $$l_1$$ and $$h_1$$, we can use the Pythagorean theorem to determine $$a$$ and proceed as in the previous paragraph. The same is true if we know $$a$$ and $$h_1$$.

That leaves two scenarios: We know $$a$$ and $$h$$, or we know $$l_1$$ and $$h_2$$.

In the first case, $$h_2 = h – h_1$$ and $$a^2 = h_1h_2 = h_1(h – h_1)$$. So $$h_1^2 – hh_1 + a^2 = 0$$. We know $$a$$ and $$h$$, so we can determine $$h_1$$ using the quadratic formula: $$h_1 = \frac{h \pm \sqrt{h^2 – 4a^2}}{2}$$. This will have two solutions (unless $$h_1 = h_2$$), one for the mirror image of the other. From here we can calculate $$h_2 = h – h_1$$, and then $$l_1$$ and $$l_2$$.

In the second case, $$hh_1 = h_1(h_1 + h_2) = l_1^2$$ and $$h_1^2 + h_2h_1 – l_1^2 = 0$$. We can then determine $$h_1$$ using the quadratic formula: $$h_1 = \frac{-h_2 \pm \sqrt{h_2^2 + 4l_1^2}}{2}$$. This will only have one positive solution, $$h_1 = \frac{-h_2 + \sqrt{h_2^2 + 4l_1^2}}{2}$$. From here, we can determine $$h = h_1 + h_2$$, and then $$l_2$$ and $$a$$.

This means that, if we have any two lengths of the six in the diagram, there is exactly one right triangle (including, in one case, a mirror image) with those measurements.

That’s really cool.