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I have seen in an old geometry textbook that the formula for the length of the angle bisector at $A$ in $\triangle\mathit{ABC}$ is \begin{equation*} m_{a} = \sqrt{bc \left[1 - \left(\frac{a}{b + c}\right)^{2}\right]} , \end{equation*} and I have seen in a much older geometry textbook that the formula for the length of the same angle bisector is \begin{equation*} m_{a} = \frac{2}{b + c} \sqrt{bcs(s - a)} . \end{equation*} ($s$ denotes the semiperimeter of the triangle.)

I did not see such formulas in Euclid's Elements. Was either formula discovered by the ancient Greeks? May someone furnish a demonstration of either of them without using Stewart's Theorem and without using the Inscribed Angle Theorem?

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    It should be noted that those 2 formulas are algebraically equivalent.$$\frac{2}{b + c} \sqrt{bcs(s - a)}$$ $$=\sqrt{bcs(s - a)\frac{4}{(b + c)^2}}$$ $$=\sqrt{bc\frac{(b+c+a)(b+c-a)}{(b + c)^2}}$$ $$=\sqrt{bc\frac{(b+c)^2-a^2}{(b + c)^2}}$$ $$=\sqrt{bc \left[1 - \left(\frac{a}{b + c}\right)^{2}\right]}$$ – Nathaniel B Jul 13 '16 at 02:02

2 Answers2

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We may prove it by avoiding the usual path, and proving other interesting things along the way.

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Let $I$ be the incenter and $I_A$ the $A$-excenter. $\widehat{IBI_A}=\widehat{ICI_A}=\frac{\pi}{2}$, hence $IBI_A C$ is a cyclic quadrilateral. By Van Obel's theorem we have $$ \frac{AI}{IL_A}=\frac{b+c}{a}, $$ hence it is enough to find $IA^2$. Let $C'$ be the symmetric of $C$ with respect to $AI$. We have: $$ AI\cdot AI_A = \text{pow}_A\left(\Gamma_{I_A B C}\right)=AB\cdot AC'=bc.$$ The problem boils down to finding $II_A=AI_A-AI$, i.e. the diameter of the circumcircle of $I_A BC$. The midpoint of $II_A$ is also the midpoint of the $BC$-arc in the circumcircle of $BCI_A$, hence the previous diameter just depends on the length of $BC$, i.e. $a$, and the angle $\widehat{BNC}=\pi-\widehat{A}$. Putting everything together, we get that $IA$ is a root of

$$ x^2+x\frac{a}{\cos\frac{A}{2}}-bc, $$

but $IA\cos\frac{A}{2}=\frac{b+c-a}{2}$ and $$ \frac{a}{\cos^2\frac{A}{2}}=\frac{2a}{1+\cos A}=\frac{4abc}{(b+c-a)(b+c+a)},$$ hence:

$$ IA^2 = bc-\frac{2abc}{a+b+c} = \color{red}{bc-4rR}.$$

Now we may exploit the parallel axis theorem to compute the unpleasant squared distance $IG^2$ in a very slick way:

$$\begin{eqnarray*} \sum_{cyc}IA^2 = ab+ac+bc-12rR &=&3IG^2+GA^2+GB^2+GC^2\\ &=& 3IG^2+\frac{3}{4}(a^2+b^2+c^2)\end{eqnarray*}$$ leads to:

$$ IG^2 = ab+ac+bc-\frac{2abc}{a+b+c}-\frac{a^2+b^2+c^2}{4}$$

and we may prove $IO^2=R^2-2Rr$ in a similar way.

Jack D'Aurizio
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  • What is $\mathrm{pow}{A}(\Gamma{I_{A}BC})$? Please provide a link to the formula you quoted. –  Jul 13 '16 at 20:15
  • @Adelyn: it is the power of the point $A$ with respect to the circumcircle of $I_A BC$ : http://www.cut-the-knot.org/pythagoras/PPower.shtml – Jack D'Aurizio Jul 13 '16 at 20:25
  • I have heard of that phrase with respect to a trapezoid but not with respect to a circle. Thanks for clarifying. –  Jul 14 '16 at 12:32
  • @Adelyn: I never heard the phrase the power of a point with respect to a trapezoid. What is it, since trapezoids are not always cyclic quadrilaterals? – Jack D'Aurizio Jul 14 '16 at 12:44
  • I guess that I can't be sure that I have heard of this phrase as it pertains to trapezoids. I thought that I had seen in about five months ago in the book Mathematical byways in Ayling, Beeling, and Ceiling by Hugh ApSimon when I was looking at formulas associated with trapezoids. I just looked in the book and the phrase wasn't there. I looked in the articles by Martin Josefsson, and the phrase wasn't there. I looked in the article "A Fascinating Application of Steiner's Theorem for Trapezium," and the phrase wasn't there. It seems that I made a mistake. –  Jul 14 '16 at 21:29
  • I am fairly sure what I meant by the phrase was the vertex of the triangle constructed in Steiner's Theorem. If the lines through the legs of a trapezoid intersect at P, the line through P and the intersection of the two diagonals of the trapezoid bisect the two bases of the trapezoid. I associated P with "the power of the point (of the trapezoid)." I apologize for the confusion. –  Jul 14 '16 at 21:30
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Here is a proof of the formula:

Let $AD$ be the angle bisector at $A$ (where $D \in BC$).

The area of $\triangle ABC$ is equal to the area of $\triangle ABD$ $+$ the area of $\triangle ACD$: $$\frac{1}{2} bc \sin\alpha=\frac{1}{2} b m_a \sin(\alpha/2) + \frac{1}{2} c m_a \sin(\alpha/2)$$ $$m_a = \frac{2bc}{b+c}\cos(\alpha/2)$$ We can compute $\cos(\alpha/2)$ in terms of $a,b,c$ by using the cosine rule: $$ 2\cos^2(\alpha/2) - 1=\cos\alpha=\frac{b^2 + c^2 - a^2}{2bc}$$ $$\cos^2(\alpha/2) = \frac{(b+c)^2 - a^2}{4bc} = \frac{s(s-a)}{bc} $$

Hence, we get: $$m_a = \sqrt{bc\left[ 1 - \left(\frac{a}{b+c}\right)^2\right]} = \frac{2}{b+c}\sqrt{bcs(s-a)}$$

W. W. Wilson
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  • This appears to be the elementary demonstration that I was requesting. You furnished a polished presentation of it. –  Jul 13 '16 at 20:19
  • That is a clever demonstration! –  Jul 15 '16 at 18:05