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### A Triangle Vertex Bisection and Its Trio of New Lengths

```Date: 06/08/2012 at 11:28:37
From: Saurabh
Subject: Length of angle bisector in a triangle

In terms of its sides and trigonometric ratios, what will be the length of
an angle bisector in any triangle having sides a, b, c and angles A, B, C?

The most difficult part is finding the lengths in which the angle bisector
divides the opposite side. Even if I assume them, and apply the law of
cosines, simplification is very difficult.

```

```
Date: 06/08/2012 at 13:08:30
From: Doctor Peterson
Subject: Re: Length of angle bisector in a triangle

Hi, Saurabh.

See the following page, which gives a formula in terms of the side lengths
only:

http://mathworld.wolfram.com/AngleBisector.html

I am not familiar with the formula given there for the length of the
bisector; but the theorem on how it divides the opposite side is well
known and easy to prove:

http://mathworld.wolfram.com/AngleBisectorTheorem.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 06/09/2012 at 04:43:42
From: Saurabh
Subject: Length of angle bisector in a triangle

Thank you very much. The formula is useful. But what is the proof?

```

```
Date: 06/09/2012 at 11:04:19
From: Doctor Peterson
Subject: Re: Length of angle bisector in a triangle

Hi, Saurabh.

Are you asking for the proof of the formula I said I was unfamiliar with,
or the proof I said was easy? I'll answer the latter, then see what I can
come up with for the former.

The Angle Bisector Theorem says

The angle bisector of an angle in a triangle divides the opposite
side in the same ratio as the sides adjacent to the angle.

Before writing up the proof myself, I just searched our site and found it
has already been done:

The Angle Bisector and Equal Side Ratios
http://mathforum.org/library/drmath/view/55014.html

My preferred way to express this, using Doctor Santu's picture, is that
the areas of two triangles with the same altitude are proportional to
their bases. This ratio of ABD:CBD is, on one hand, equal to the ratio of
bases AD:CD, with the red segment as their common altitude; and, on the
other hand, to AB:BC, with the blue segments as their common altitude.
Therefore,

a:b = x:y

This is the form in which the theorem was stated above.

Now, let's look at the other formula:

The length t1 of the bisector A1T1 of angle A1 in the above
triangle A1A2A3 is given by ...

t1^2 = a2a3[1 - a1^2/(a2 + a3)^2],

... where ti = AiTi and ai = AjAk.

To make things easier to write, let's set

A1 = B
A2 = A
A3 = C
T1 = D
a1 = AC = b
a2 = BC = a
a3 = AB = c

With this different labeling, we have

B
/  \
/  /  \
c  /        \ a
/     /     \
/        t     \
/        /        \
/     x        y     \
A-----------D-----------C
b

This formula says

The length t of the bisector BD of angle B in the above
triangle BAC is given by

t^2 = ac[1 - b^2/(a + c)^2]

To prove this, we know the following:

x + y = b
x/y = c/a

This is equivalent to

ax - cy = 0

Solving for x and y,

cx + cy = bc
ax - cy = 0
-------------
(a + c)x = bc

x = bc/(a + c)
y = ab/(a + c)

We can also apply the law of cosines to triangles ADB and BDC, as you
suggested. If we call cos(ADB) "q," then cos(BDC) = -cos(ADB) = -q.

So we have

c^2 = x^2 + t^2 - 2xtq
a^2 = y^2 + t^2 + 2ytq

Multiplying the first by y and the second by x, then adding to eliminate
q, we get

xa^2 + yc^2 = xy^2 + x^2y + (x + y)t^2

Solving for t^2,

t^2 = [xa^2 + yc^2 - (x + y)xy]/(x + y)

t^2 = [xa^2 + yc^2 - bxy]/b

Plugging in expressions for x and y,

t^2 = [a^2bc/(a + c) + c^2ab/(a + c) - b(bc/(a + c))(ab/(a + c))]/b
= [(a)abc/(a + c) + (c)abc/(a + c) - ab^3c/(a + c)^2]/b
= [(a + c)abc/(a + c) - ab^3c/(a + c)^2]/b
= (a + c)ac/(a + c) - ab^2c/(a + c)^2
= ac[1 - b^2/(a + c)^2]

There's the formula!

It sounds like your attempt was essentially the same, but less efficient
and more error-prone.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 06/10/2012 at 04:08:50
From: Saurabh
Subject: Thank you (Length of angle bisector in a triangle)

Thank you very much!!!
```
Associated Topics:
High School Triangles and Other Polygons
High School Trigonometry

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