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### The Angle Bisector and Equal Side Ratios

```
Date: 05/17/98 at 17:17:07
From: Jason Yerep
Subject: Geometry Proof

Given: Triangle ABC
Angle bisector BD

B
/|\
x   / | \  y
/  |  \
/   |   \
/    |    \
/     |     \
A/      |      \C
---------------
a    D   b

Prove:

x     y
--- = ---
a     b

where x = side AB, a = segment AD, y = side BC, and b = segment DC.
```

```
Date: 05/19/98 at 10:06:06
From: Doctor Santu
Subject: Re: Geometry Proof

Dear Jason:

This problem can be solved by comparing the areas of the two
triangles. The crucial idea is that angle bisectors are equally
distant from both sides of the angle. In other words, if you have an
angle, and you draw the angle bisector, no matter which point on the
bisector you pick, it will be the same distance from the two lines
of the original angle. This is the main idea in the solution I came
up with.

Below is a figure drawn so that it doesn't look like an isosceles
triangle.

See the blue lines? They're perpendiculars from the point D to the two
sides of the top angle. Since D is on the angle bisector, it's the
same (perpendicular) distance from the two sides BA and BC. How do we
know this? The two triangles with BD as a common side and containing
the blue lines are congruent (SAA) so the blue lines, being
corresponding parts, must be congruent.

Now forget the blue lines for the moment -- imagine that they're not
there. Ignore the red line, too, and what you have left is the
original triangle you drew, with the line down the middle. Let's call
the areas of the two triangles A1 and A2. Then:

A1 = (1/2)*base*height = (1/2)*a*h

where h is the altitude of the left triangle, the length of the red
line. Also:

A2 = (1/2)*b*h

since the triangles have the same altitude, from the way their bases
are collinear, and have a vertex in common.

But you can figure the two areas in different ways, too. The triangle
on the left can be turned around so that the blue line on the left
is an altitude. Now its area is:

A1 = (1/2)*x*k

where k is the length of the blue line.

In the same way, the area of the triangle on the right is:

A2 = (1/2)*y*k

So a*h = x*k and b*h = y*k. A simple calculation shows that:

x/y = a/b

Good luck,

-Doctor Santu,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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