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Essay text:
We get the corner:
We have .
Let’s put our corner in the zero vector point:
Then A=(0,0,0), D=(a,0,0), C=(0,b,0), B=(0,0,c).
Similarly,
The magnitude of cross product of two vector gives us the area of the parallelogram. So, the area of the triangle BCD is one half from the magnitude of cross product of vectors BD and BC.
Let
p=Area of ?ABC
r=Area of ?ACD
q=Area of ?ABD
s=Area of ?BCD
Then we see that
But we will try to find something looking like
Let’s take an isosceles triangle ABC, where AB=AC=r, take a point D on BC and take BD=p, CD=q. Without loss of generality take p?q. Let AD=s...
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