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alright, thanks. I think I understand this now.
how did you go from the step: BP/PA = QC/AQ (which is just similar triangle ratios) to
(BP/PA)(AQ/QC) = 1?
Thanks
1) Prove: if p and q are points on sides AB and AC, respectively, of trianlge ABC so that PQ is parallel to BC and if X is the point of intersection of BQ and CP then AX goes through the midpoint of BC.
The hint is to use Ceva's Theorem. I'm not sure how to start. Any help is appreciated.
The triangle that you draw has a length of ?(a²+b²), where a and b are the lengths of the other two sides.
How do you get this? Thanks.
It says to construct an isosceles right triangle whose area is equal to the sum of the areas of the two given isosceles right triangles.
And there are two given isosceles right triangles, one smaller, one larger.
The teacher hinted to do some calculation but of course, the straight edge is unmarked, so i'm not sure how to even start the hint.
Any help is appreciated
Thanks
The 'solution' / hint in the back of the book says
if a > 0 and 1/a <= 0
then 1= a*(1/a) <= a*0 =0.
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Can anyone decipher what is going on here? Is this a contradiction proof? Is it assuming a>0 and 1/a<=0 (which would be not 1/a >0)
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So does this sound like a contradiction proof,
assume a>0 and 1/a <= 0.
then 1/a = [a(1/a)] / a <= 0.
so a*(1/a) <= 0. (multiplying both sides by a)
but this contradicts our assumption that a > 0. (since a* 1/a is assumed greater than 0.)
Therefore, 1/a > 0.
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Thanks for feedback
I'm having a hard time proving this.
prove when a is a real number if a > 0 then 1/a > 0.
It seems so obvious, but i don't see a way to write a formal proof. Maybe by contradiction, but I still can't see how to write it.
Any help is appreciated
Thanks
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