Bisecting a triangle

Geometry Level 2

In triangle $ABC$ , $\angle ABC = 30^\circ , \angle ACB = 60^\circ$ . $D$ is a point in triangle $ABC$ such that $DB$ and $DC$ bisect angles $ABC$ and $ACB$ respectively. What is the measure (in degrees) of $\angle BDC$ ?

The answer is 135.

4 solutions

Jing Ni Ng
Aug 14, 2013

Since DB bisects (cuts into two equal parts) ∠ABC, and DC bisects ∠ACB, ∠DBC and ∠DCB in the new triangle, that is, triangle BDC, will be 15∘ and 30∘. Thus, take 180∘ - (15∘+30∘) = 135∘

Kenneth Cajefe
Aug 12, 2013

180-(30/2 + 60/2)

=180-(15+30)

=180-45

=135

Moderator note:

You should explain your steps, so that other people who look at it can easily understand what you are doing.

thank you!

Rashi Ranjan - 7 years, 10 months ago

David Kroell
Aug 16, 2013

If it bisects all angles then all other angles inside a New Triangle must be divided by two.

30/2 = 15 , 60/2 = 30
Since every triangles interior angle sum is 180 degrees we have to do 180 - 15 - 30 .
The answer to the equation is: 135 Degrees

Great explanation I did this the same way

Alex Hepworth - 4 years, 11 months ago

Ayon Pal
Aug 11, 2013

$\angle DBC = 30^o/2 = 15^o$

$\angle DCB = 60^o/2 = 30^o$

then, $\angle BDC = 180^o - (15^o + 30^o) = 135^o$

Moderator note:

Do you know how to show that $DA$ bisects angle $BAC$ ? This is known as the concurrency of the angle bisectors.

For any equal sided polygon the sum of internal angles is (n-2)*180 degrees. Bisecting the given angles creates another triangle with internal angles 15 and 30 therefore 180-15-30=135 degrees is the remaining angle.

Ted van Ryn - 7 years, 10 months ago

Bisecting a triangle

The answer is 135.

4 solutions

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