What is the proof for triangle ABC with known sides and angle relationships?

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Blissfulpain
ok, I'm in first year coolege and we're doing some review from high school work.. but i can't for the life of me remember how to do one of these questions and i can't seem to get past this one part or at least figure out how to prove that it is what i think it is.

it's going to be a bit trickier, because i can't insert a picture, but i guess that isn't of the most importance anyways. alright

you have triangle ABC.
side AB=4 units
side AC=7 units
the line CD extends for 3 units towards point B, the distance between point B an D is unknown and is what needs to be found
the Angle BCD is = angle DAC

here's what I've tried to do so far...

i've tried to fold the 2 triangles in half across the line AD, which lines up the two sides AB and AC, with the line AC extending 3 units futher then line AB. I'm still trying to follow this train of reasoning.. but so far with little success, i don't really know what to do next, but i think it helps somehow : )

next...

I tried extending the common side between the two triangles, line AD, far enough so that i can draw a new line, 7 units long, from point C, i'll call the intersection point E. The problem arises when i can't prove that the new triangle DEC is the same as triangle ADC. I CAN prove that it is proportional to triangle ABD, but that doesn't help me, because i only know that the side CE is 7 units, the Side DC is 3 units, and i don't know if i can prove that the angle DEC is equal to the angle DAC... which means i can't prove that the line DE is a continuation of the line AD...

hence I'm lost.

thx for the help in advance.. hey, kewl... just found the attacthment thing... i'll go draw a picture and then post :)
 

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Hint: angle bisector theorem.

Hope that helps!
 
nope, doesn't help... i looked it up on the net, and got this "The angle bisector of an angle in a triangle divides the opposite side in the same ratio as the sides adjacent to the angle."

so the angle bisector is line AD, and the adjacent and opposiye side are? i can't assume ANY right angles... and all the other stuff i found on this theorem had something to do with area of a triangle, which i can't do either, because i don't have enough info to find that yet...

a bit more of a hint is needed :confused:
 
Blissfulpain said:
nope, doesn't help... i looked it up on the net, and got this "The angle bisector of an angle in a triangle divides the opposite side in the same ratio as the sides adjacent to the angle."

so the angle bisector is line AD, and the adjacent and opposiye side are? i can't assume ANY right angles... and all the other stuff i found on this theorem had something to do with area of a triangle, which i can't do either, because i don't have enough info to find that yet...

a bit more of a hint is needed :confused:

No right angles required, merely an understanding of the words adjacent and opposite.

Let me paraphrase the theorem for the triangle you've got :

Simply, it says that CD/BD = AC/AB.

Now go back and understand what the theorem is trying to say.
 
so, 3/x=4/7
3/x=1.75
x=1.75*3
x=5.25
huh, i had gotten that number before... just couldn't prove that it was the right number... now i can with that theorem

thx a lot guys :)))))
 
oops... i mean 3/7=x/4
so 3/7=.429
.429x4=1.7 units

therefore BD=1.7
 

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