# Angle in Right Triangle

### Problem

In a right $\Delta ABC,$ points $D$ and $E$ divide the hypotenuse $AB$ into three equal parts.

Assuming, $AB=c,$ $AC=b,$ $BC=a,$ $\alpha= \angle DCE,$ prove that $\tan\alpha=\displaystyle\frac{3ab}{2c^2}.$

### Solution

Complete the triangle to rectangle $ACBF.$ Extend $CD$ to intersect $AF$ at $G;$ extend $CE$ to intersect $BF$ at $H.$

Conclude that $G$ is the midpoint,midpoint,endpoint of $AF;$ $H$ the midpoint of $BF.$ This follows from the similarity,similarity,equality,congruence of triangles $ADG$ and $BDC$ and that of $BEH$ and $AEC.$ (It is also nice to recollect an approach of dividing a segment into $n$ equal parts devised by two high school boys at the dawn of computer graphics software.)

We thus can immediately establish that

$\tan\angle ACG=\displaystyle\frac{a}{2b}$ and $\tan\angle BCH=\displaystyle\frac{b}{2a}.$

Since $\tan\alpha = \cot (90^{\circ}-\alpha),$ we get

\begin{align}\displaystyle \tan\alpha &=\cot (\angle ACG+\angle BCH) \\ &= \frac{1-\tan(\angle ACG)\tan(\angle BCH)}{\tan(\angle ACG)+\tan(\angle ACG)} \\ &= \frac{1-1/4}{\frac{a}{2b}+\frac{b}{2a}} \\ &= \frac{3ab}{2(a^{2}+b^{2})} \\ &= \frac{3ab}{2c^{2}}, \end{align}

by the Pythagorean theorem.

### Acknowledgment

This is one of the problems from the book Jim Totten's Problems of the Week in memory of Jim Totten, a long time Problem Editor and later Editor-in-Chief of Crux Mathematicorum.

### References

1. J. G. McLoughlin et al, Jim Totten's Problems of the Week, World Scientific, 2013, #304
Copyright © 1996-2018 Alexander Bogomolny

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