Geometric mean for angle bisectors

The three traditional triangle cevians are medians, altitudes, and angle bisectors. Of these three, only the altitude is usually considered in the geometric mean theorem. However, the median also has an associated, isolated geometric mean theorem (example 1). Since what is good for the goose is also good for the angle bisector, we thought it good to present a geometric mean theorem for this third triangle cevian.

Let a,b,c be the side-lengths of \triangle ABC, and let l be the length of the bisector of \angle C, as shown below:

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Then l is the geometric mean of m and n if, and only if:

(1)   \begin{equation*} \begin{split} (a+b)^2&=2c^2 \end{split} \end{equation*}

We’ll see the simplifications that result when equation (1) restricts to right triangles

(2)   \begin{equation*} \begin{split} c^2&=a^2+b^2 \end{split} \end{equation*}

and the simplifications that result when (1) is considered in the context of triangles satisfying

(3)   \begin{equation*} (a^2-b^2)^2=(ac)^2+(cb)^2 \end{equation*}

Let’s set the ball rolling with what a right triangle enjoys solely.

PROVE that a median is the geometric mean of the two segments into which it divides the opposite side if, and only if, the parent triangle is a right triangle.

Suppose that \triangle ABC is a right triangle with \angle C=90^{\circ}. Then the median from vertex C is half of the length of the hypotenuse: m_c=\frac{1}{2}c and so m_c is the geometric mean of \frac{1}{2}c and \frac{1}{2}c. On the other hand, if a triangle has the property that one of its medians is the geometric mean of the two segments on the opposite side, for example, m_c^2=\frac{1}{2}c\times \frac{1}{2}c, we show that the triangle is a right triangle. Indeed:

    \begin{equation*} \begin{split} m_c^2&=\frac{1}{2}c\times \frac{1}{2}c\\ \frac{2a^2+2b^2-c^2}{4}&=\frac{1}{4}c^2\\ 2a^2+2b^2&=2c^2\\ \therefore a^2+b^2&=c^2\\ \end{split} \end{equation*}

So the triangle is a right triangle. Since all medians are internal, this property is unique to right triangles.

Formula derivation

Our main result is example 4, for which we may need the next two examples.

In the diagram below, CD is the bisector of \angle C. PROVE that m=\frac{b}{a+b}c and n=\frac{a}{a+b}c.

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We first have that m+n=c. Also:

    \[\angle ADC + \angle CDB=180^{\circ}\implies \sin \angle ADC=\sin \angle CDB\]

The sine rule applied to \triangles ADC and CDB then gives:

    \begin{equation*} \begin{split} \frac{m}{\sin\theta}&=\frac{b}{\sin \angle ADC}\\ \frac{n}{\sin\theta}&=\frac{a}{\sin \angle CDB}\\ \therefore \frac{m}{n}&=\frac{b}{a}\\ m&=\frac{b}{a}n\\ c-n&=\frac{b}{a}n\\ c&=\left(\frac{b}{a}+1\right)n\\ c&=\frac{a+b}{a}n\\ \therefore n&=\frac{a}{a+b}c\\ \therefore m&=\frac{b}{a+b}c\\ \end{split} \end{equation*}

Basically, m:n=b:a. This is very useful in finding the coordinates of the foot of an angle bisector, which in turn is useful in finding the coordinates of the incenter of a triangle.

PROVE that l^2=ab\left(1-\left(\frac{c}{a+b}\right)^2\right), where l is the length of the bisector of \angle C.

Let’s refer to the diagram in example 2 again:

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and then apply Stewart’s theorem to \triangle ABC to obtain

    \[a^2m+b^2n=c(l^2+mn)\implies l^2=\frac{a^2m+b^2n}{c}-mn\]

We now use the fact that m=\frac{b}{a+b}c and n=\frac{a}{a+b}c from example 2.

    \begin{equation*} \begin{split} l^2&=\frac{a^2m+b^2n}{c}-mn\\ &=\frac{a^2\times \frac{b}{a+b}c+b^2\times \frac{a}{a+b}c}{c}-\frac{b}{a+b}c\times \frac{a}{a+b}c\\ &=ab-\frac{abc^2}{(a+b)^2}\\ &=ab\left(1-\frac{c^2}{(a+b)^2}\right) \end{split} \end{equation*}

An alternative procedure is to use area arguments as well as the cosine formula and double angle identities for \sin 2\theta and \cos 2\theta. This way one avoids going through Stewart’s theorem and an explicit determination of m and n.

In the diagram below, PROVE that the length l of the bisector of \angle C is the geometric mean of the segments AD and DB if, and only if, (a+b)^2=2c^2.

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First suppose that l is the geometric mean of m and n. Then l^2=mn.

    \begin{equation*} \begin{split} \implies ab\left(1-\left(\frac{c}{a+b}\right)^2\right)&=\frac{a}{a+b}c\times \frac{b}{a+b}c\\ 1-\frac{c^2}{(a+b)^2}&=\frac{c^2}{(a+b)^2}\\ 1&=2\frac{c^2}{(a+b)^2}\\ \therefore (a+b)^2&=2c^2\\ \end{split} \end{equation*}

Conversely, suppose that (a+b)^2=2c^2.

    \begin{equation*} \begin{split} l^2&= ab\left(1-\left(\frac{c}{a+b}\right)^2\right)\\ &=ab\left(1-\frac{c^2}{2c^2}\right)\\ &=\frac{1}{2}(ab)\\ mn&=\frac{b}{a+b}c\times \frac{a}{a+b}c\\ &=\frac{abc^2}{(a+b)^2}\\ &=\frac{abc^2}{2c^2}\\ &=\frac{1}{2}ab\\ \therefore l^2&=mn \end{split} \end{equation*}

Notice how we obtain another equivalence: l^2=mn\iff (\sqrt{2}l)^2=ab. Translate into words: we killed two birds with one stone.

For a quick example consider triangle ABC with vertices at A(0,0), B\left(6\sqrt{2},0\right), and C\left(2\sqrt{2},\sqrt{17}\right), shown below:

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The foot D of the bisector of \angle C has coordinates \left(\frac{5}{2}\sqrt{2},0\right):=D. By the distance formula:

    \[l^2=\left(\frac{5}{2}\sqrt{2}-2\sqrt{2}\right)^2+\left(0-\sqrt{17}\right)^2=\frac{35}{2}.\]

Also:

    \[m=\frac{5}{2}\sqrt{2},~n=\frac{7}{2}\sqrt{2}\implies mn=\frac{35}{2}=l^2.\]

Thus, CD is the geometric mean of AD and DB.

Fifteen degrees

We now restrict equation (1) to right triangles and to triangles satisfying equation (3) . It turns out that the interior angles are all multiples of 15^{\circ}.

If the bisector of \angle C=90^{\circ} in a right triangle satisfies equation (1), PROVE that the triangle is isosceles.

Using the fact that a^2+b^2=c^2 and (a+b)^2=2c^2, we have:

    \[(a+b)^2=2(a^2+b^2)\implies (a-b)^2=0\implies a=b\]

The angles of the triangle will then be 45^{\circ}, 45^{\circ}, 90^{\circ}. All multiples of 15^{\circ}.

If the bisector of the apex \angle C in an isosceles triangle satisfies equation (1), PROVE that the triangle is a right triangle.

In this case a=b. Together with equation (1), we have

    \[(a+b)^2=2c^2\implies (2a)^2=2c^2\implies c^2=2a^2=a^2+b^2\]

Alternatively, the bisector is a median in this case, and so the conclusion follows from example 1 and example 4.

Suppose that \triangle ABC satisfies equation (3). If the bisector of \angle C satisfies equation (1), PROVE that a^2-4ab+b^2=0.

We have: (a+b)^2=2c^2\& (a^2-b^2)^2 =c^2(a^2+b^2). Isolate c^2 and simplify:

    \[(a+b)^2=2\left(\frac{(a^2-b^2)^2}{a^2+b^2}\right)\implies a^2+b^2=2(a-b)^2\implies \scriptstyle a^2-4ab+b^2=0\]

Suppose that \triangle ABC satisfies equation (3). If the bisector of \angle C satisfies equation (1), PROVE that \angle A=15^{\circ}, \angle B=105^{\circ}, \angle C=60^{\circ}.

Since \cos C=\frac{2ab}{a^2+b^2} (see here), we have (by example 7 above):

    \[\cos C=\frac{2ab}{4ab}=\frac{1}{2}\implies \angle C=60^{\circ}\]

Also, \cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)} and \cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)} yield \angle A=105^{\circ} and \angle B=15^{\circ} (or \angle A=15^{\circ} and \angle B=105^{\circ}).

Further developments

Here are two out of a few more consequences.

Let a,b,c be the side-lengths of \triangle ABC. If equations (1) and (3) are satisfied, PROVE that the circumradius R is the geometric mean of a and b.

We have a^2+b^2=4R^2 and a^2-4ab+b^2=0. Eliminate a^2+b^2 to obtain R^2=ab.

If \triangle ABC satisfies equations (1) and (3), PROVE that the squared lengths h_C^2,l^2,R^2 form a geometric progression. Here, l is the length of the bisector of \angle C, h_C is the length of the altitude from vertex C, and R is the circumradius.

In example 4 we had l^2=\frac{1}{2}(ab); in example 9 above we had R^2=ab. Now let’s calculate h_C:

    \[h_C=\frac{ab}{2R} =\frac{ab}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{4ab}}\implies h_C^2=\frac{1}{4}(ab)\]

We see that h_C^2,l^2,R^2 form a geometric progression with common ratio r=2.

Takeaway

In triangle ABC, let D be the foot of the bisector of \angle C. Then the two statements below are equivalent:

  • CD is the geometric mean of AD and DB
  • \sqrt{2} CD is the geometric mean of AC and CB.

Further, for any triangle ABC whose side-lengths a,b,c satisfy equation (1), the following two statements are equivalent:

  • \triangle ABC is a right triangle
  • a^2-2ab+b^2=0.

Finally, for any triangle ABC whose side-lengths a,b,c satisfy equation (1), the two statements below are equivalent:

Tasks

  1. Using \triangle ABC with vertices at A(-6,0), B(4,4), C(2,4), verify that:
    • its incenter is I\left(-3+\sqrt{5},-1+\sqrt{5}\right).
    • the ratio of the coordinates (x,y) of the incenter is golden; that is, it satisfies x^2-xy-y^2=0\implies \left(\frac{x}{y}\right)^2-\frac{x}{y}-1=0.
    • the slope of the bisector of \angle B is -\frac{1+\sqrt{5}}{2}, the negative of the golden ratio.
  2. Let a,b,c be the side-lengths of triangle ABC.
    • PROVE that the harmonic mean of \frac{ac}{a+b} and \frac{bc}{a+b} is \frac{2abc}{(a+b)^2}.
    • Deduce that the bisector of \angle C is the harmonic mean of the two segments on the opposite, if (a+b)^2=c^2+2c.
  3. Suppose that the side-lengths a,b,c of triangle ABC satisfy equation (3), namely (a^2-b^2)^2=(ac)^2+(cb)^2.
    • PROVE that the length l of the bisector of \angle C satisfies l^2=\frac{2a^2b^2}{a^2+b^2}.
    • Deduce that l^2 is the harmonic mean of a^2 and b^2.
  4. Suppose that \triangle ABC satisfies the usual Pythagorean identity c^2=a^2+b^2.
    • If the bisector of \angle A satisfies equation (1) adapted as (b+c)^2=2a^2, PROVE that a^2=8b^2.
    • Under the above, deduce that \sin A=\frac{2\sqrt{2}}{3}, \sin B=\frac{1}{3}.
  5. In \triangle ABC, let the side-lengths be a,b,c, and let R,h_C,l denote the circumradius, altitude from C, and the length of the bisector of \angle C. If \triangle ABC satisfies equation (3), PROVE that the squared lengths h_C^2,l^2,R^2 form a geometric progression with common ratio r=2.
  6. In \triangle ABC, let R,h_C,m_C denote the circumradius, altitude from C, and the median from C. If \triangle ABC satisfies equations (1) and (3), PROVE that m_C^2=h_C^2+R^2.
    (Basically a right triangle can be formed with the median m_C as the hypotenuse.)