# Similar and congruent triangles

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.



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

• 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. Jan 7, 2021 at 12:12

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, $$\triangle ADC \sim \triangle BEC$$
• Yes that seems right. But, is $A'B'C'$ given similar to $ABC$ ? Jan 7, 2021 at 13:15
• 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! Jan 7, 2021 at 13:54