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Let $ABC$ be a triangle and $h_a,h_b,h_c$ the length of the heights.

(a) Find similar triangles at which it holds that $ah_a=bh_b$.

(b) Use (a) to show : Is $A'B'C'$ a second triangle and $h_a,h_b,h_c$ are again the length of the heights of this triangle, so these triangles are congruent.

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For (a) do we have to give two specific triangles, i.e. give the negth of the sides? Or what exactly is asked?

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  • $\begingroup$ I don't get (a). The area of the triangle is always 1/2 times height time side length. This quantity is independent of the side chosen. $\endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Jan 7, 2021 at 12:12

1 Answer 1

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For part (a), yes, we have to find two similar triangles for which $ah_a=bh_b$ holds.

Draw $h_a$ and $h_b$ inside $\triangle ABC$. $h_a \cap BC \in D$. $h_b \cap CA \in E$. With simple angle chasing, $$\triangle ADC \sim \triangle BEC$$

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    $\begingroup$ Yes that is absolutely correct. $\endgroup$
    – cosmo5
    Jan 7, 2021 at 13:09
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    $\begingroup$ Yes that seems right. But, is $A'B'C'$ given similar to $ABC$ ? $\endgroup$
    – cosmo5
    Jan 7, 2021 at 13:15
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    $\begingroup$ @MaryStar Indeed, we must show second triangle has same angles and proportional sides for them to be congruent. I'm not sure if this results holds, that given three altitudes, a unique triangle exists. I'd like to see others' take on this too. $\endgroup$
    – cosmo5
    Jan 7, 2021 at 13:34
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    $\begingroup$ I see. One similarity is not enough. For example, $3,4,5$ and $3,4,6$ triangles are not similar. However taking pairwise altitudes, we do get that $a/a'=b/b'=c/c'$ from which it follows. Nice question! $\endgroup$
    – cosmo5
    Jan 7, 2021 at 13:54
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    $\begingroup$ Yes, you're right. @MaryStar $\endgroup$
    – cosmo5
    Jan 7, 2021 at 15:54

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