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why does the triangle angle bisector theorem work?
This is for an extra credit assignment that i need greatly! if i can explain with a picture how it works it will be done corrrectly... any idea is helpful!!!!
- 1 decade agoFavorite Answer
Can you follow http://planetmath.org/?op=getobj&from=objects&id=6... ?
I'll go through it a little more slowly...
First, add a height line in your drawing perpendicular to BC up to A, because that'll make life easier and makes it look less like you copied it from the Web.
First step: The ratio of areas is the same as the ratio of the line segments.
Second step, we're finding the areas in a different way. Remember sin = opp/hyp?
- For BAP, the hypotenuse is BA, and the base is AP, and BA*sin(BAP) gives us another height from AP to B, perpendicular to AP, so area is (1/2)bh because h=BA*sin(BAP)...
- For CAP, the hypotenuse is CA, the base is AP, and we get a new height that, using this particular triangle, I would draw by extending AP until you can draw a line perpendicular to AP up to C.
So, now we've got one formula for the ratio of the areas of the triangle (step 1) relating BP and PC to each other, and then we've got two separate equations for the area of each triangle, so we're going to pop those 2nd 2 equations into the areas of the first equation, and we've got some cancellation to do (AP/2 cancels out), leaving BP/PC = BA(sinBAP) / CA(sinCAP)... and when you bisect CAB, BAP=CAP, and THEY cancel out, so BP/PC=BA/CA.