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Home » Triangles » Angle Bisector Theorem

Angle Bisector Theorem

Last updated: Mar 27, 2021 by Ido Sarig · This website generates income via ads and uses cookies · Terms of use · Privacy policy

In today's lesson, we will show a straightforward way of proving the Angle Bisector Theorem.

The angle bisector is a line that divides an angle into two equal halves, each with the same angle measure.

The angle bisector theorem states that in a triangle, the angle bisector partitions the opposite side of the triangle into two segments, with a ratio that is the same as the ratio between the two sides forming the angle it bisects:

angle bisector

If ∠BAD≅ ∠CAD, then |BD|/|DC|=|AB|/|AC|

This is another useful tool in problems that require you to compare lengths of different line segments. [The others being similar triangles, triangles with the same height or base, or the intercept theorem].

Problem

AD is the angle bisector of angle ∠BAC in triangle △ABC (∠BAD≅ ∠CAD). Show that |BD|/|DC|=|AB|/|AC|

Strategy

Many proofs of this theorem use trigonometry and the law of sines. But here, we will provide a proof that does not rely on such advanced knowledge.

Instead, we will use one of the other tools already in our pocket for comparing ratios of line segments- triangles with the same height.

Let's look at the two triangles formed by the angle bisector, △ABD and △ADC. Both have the same height, h, from A:

angle bisector with height

That means that the ratio of their areas will be the same as the ratio of their bases:

AreaΔABD/ AreaADC=|BD|/|DC|

Now, as we've seen, all points on the angle bisector are equidistant from the two sides of the angle. (This is easily proven with congruent triangles and the angle-side-angle postulate). D is such a point on the angle bisector.

points on bisector are equidistant

So, if we compute at the areas of these two triangles using the height from D, which is the same for both triangles, we have:

AreaΔABD=(|AB|·h1)/2, and AreaΔACD=(|AC|·h1)/2, so

AreaΔABD/AreaΔACD=|AB|/|AC|.

But as we saw above, AreaΔABD/ AreaADC=|BD|/|DC| , so |BD|/|DC|=|AB|/|AC|

Proof

(1) AreaΔABD/ AreaΔADC=|BD|/|DC| //ratio of areas of triangles with same height is equal to the ratio of their bases
(2) AD=AD //Common side, reflexive property of equality
(3) ∠BAD≅ ∠CAD //given, AD is the angle bisector of ∠BAC
(4) m∠DEA=m∠DFA=90° //construction
(5) ∠EDA≅ ∠FDA //(3),(4), Sum of angles in a triangle
(6)△EDA≅△FDA //(2), (3), (5), Angle-Side-Angle Postulate
(7) DE=DF=h1 //Corresponding sides of congruent triangles (CPCTC)
(8) AreaΔABD/AreaΔACD=|AB|/|AC| //ratio of areas of triangles with same height is equal to the ratio of their bases
(9) |BD|/|DC|=|AB|/|AC| //(1),(8), transitive property of equality

« Radius of a Regular Polygon
Angle Bisector Equidistant Theorem »

About the Author

Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. His goal is to help you develop a better way to approach and solve geometry problems. You can contact him at [email protected]

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Welcome to Geometry Help! I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree in Management of Technology. I'm here to tell you that geometry doesn't have to be so hard! My goal with this website is to help you develop a better way to approach and solve geometry problems, even if spatial awareness is not your strongest quality. Read More…

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