angle bisectors

[kastamonu]

ABC is atriangle.
BD is an internal angle bisector.
CD is an external angle bisector.
Angle DAC is alpha.
Then what is the value of angle alpha?


I found everything except alpha and Angle ADE

 

[kastamonu]

DE DC and BC are equal forming isosceles triangles.

 

[kastamonu]

Did you get angleAED = 108?

yes

 

[kastamonu]

Good work!

Let AB=1; then BE=1 and CE=1 ; got that?
Let e = AE and a = DE = DC = BC
4 steps to get angle DAE:

1: e = SIN(36) / SIN(72)

2: a = SIN(72) / SIN(36)

3: d = SQRT[a^2 + e^2 - 2aeCOS(108)]

4: angleDAE = ASIN[aSIN(108) / d]

You should get a "nice" angle

There must be some other way.:-|

 

[kastamonu]

I assumed you were learning trigonometry...
What I gave you does the job: angleDAE = 54 (so angleADE = 18).

Is this homework? Were you given specific instructions, like no trigonemetry?

Anyway, this'll do it too:
extend line BA to F, such that FD is parallel to BC:
you now have parallelogram BCDF, with ALL sides equal,
angles C and F = 108, angles B and D = 72.

Play with that...it's your homework, not ours...

No it is not.I was just stuck with that one.

 

[kastamonu]

It is wrong. Answer must be 54. Thanks Denis.

 

[kastamonu]

I couldn't get 54 with this method.Maybe I missed something.

 

[mmm4444bot]

There must be some other way.

There are several ways.

If you're looking for something specific, you need to be specific. :cool:

 

[stapel]

I couldn't get 54 with this method.Maybe I missed something.

Until you show your work, I'm afraid we can't help you find what you missed. Sorry.

 

[kastamonu]

He'll never want to show his work to "idiots"

If we extend BC to F ,DCF becomes the external angle of the triangle DBC. BDC=X, DBC=X, DCF=2X.I found DCF=72 degrees.CDB=36 degrees.DBC=36 dg ABE=36 dg.AED =108 dg. I couldn't get ADE.