triangle proof: Show that a = b cosC + c cosB

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soccerisgreat

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Show that for any triangle, ABC, even if B or C is an obtuse angle, a=b cos C + c cos B.

I think if someone could start me, I could probably complete this. I just don't know how to get started.
 
Re: triangle proof

I'll give you first few steps:

1) Draw ABC with BC(a) as the base - A at the vertex. Assume B is >90°

2) Drop a perpendicular AP onto BC from A.[edited]

3) Now you have:

BC = CP - PB

Now continue...
 
How can I drop a perpendicular from A onto AB? A is the vertex already on AB. Or, do you want me to construct P on AB such that BC = CP - PB?
 
Okay, my apologies, but that didn't help much. I guess I know less than what I thought. I don't want the answer but I think I need a bit more help.
This is what I know.....
b^2=a^2+c^2-2ac cos B
I was thinking I could use that somehow.
 
How much is PC in terms of side AC(b) and angle ACB?

Similarly, how much is PB in terms of side AB(c) and angle ABP?

What is the relationship between the angles ABC and ABP?

Put all these together.
 
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