Foot of the altitude $K$ from vertex $C$ in triangle $ABC$ divides side $c$ in ratio $1:2$, proof inequality $3(a-b)
Triangle sketch:
Foot of the altitude $K$ divides side $c$ into ratio $1:2$, prove $3(a-b) < c$
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geometry
triangles
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0What do you mean "h" in the diagram and how does it relate to the inequality that must be proved_ – 2017-01-09
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0@DonAntonio h is height of triangle. I want to prove that if I choose any lenght for height h, 3(a-b) < c – 2017-01-09
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0@S Well, saying the segment of line with h by its side is one of the triangle's heights is a rather important data: write it down in your question! Now, what do you mean "if I choose any length for the height"? You can't do that: for a given triangle and a given side of it there is one single height to that side and it has a definite length. You can't "choose it"! – 2017-01-09
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0@DonAntonio Sorry, I expressed badly, $K$ is and intersect of height $h$ and side $c$. I want to prove, that for any triangle which has height to side $c$ while $K$ divides side $c$ into ratio $1:2$, $3(a-b)
1 Answers
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By Pythagorean's theorem:
$$a^2-h^2=(2x)^2=\frac{4c^2}{9},\quad b^2-h^2=x^2=\frac{c^2}{9}.$$
Hence
$$a^2-b^2=\frac{c^2}{3}\quad\Rightarrow 3(a-b)=c\cdot\frac{c}{a+b}