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Now, the question is fairly specific, $ABC$ is a triangle, and $AD$ is an altitude. Midpoints of the sides being $A'$ for $BC$,$B'$ for $CA$ and $C$' for $AB$.

Edit: By similarity I proved the <B'DA'=<BCA. Now the catch here is, this was done on the assumption that D lies on the circumcircle of A'B'C'.

So I think the MAIN QUESTION: Prove the circumcircle of A'B'C' also passes through the feet of all the perpendiculars of ABC.

Lalit Tolani
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A Guru
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  • Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. – José Carlos Santos Aug 04 '21 at 13:49
  • Please Show your attempts. And if possible write your questions separately – Lion Heart Aug 04 '21 at 13:50
  • There is some well known $9$-points-circle, alias Euler circle, https://en.wikipedia.org/wiki/Nine-point_circle, the three mid points, and the three feet of the heights are lying on it... – dan_fulea Aug 04 '21 at 15:21

2 Answers2

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enter image description here

As you can see in figure b Welcome to MES. your assumption that D lies on circumcircle of A'B'C' is correct. this circle is called nine point circle. But $\angle B'DA'<\angle BCA$. there may be an error in statement.

sirous
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The initial problem is not correct. $\angle B'DA'$ is not equal to $\angle BCA$ when $AC\leq AB$. There should be additional condition $AC> AB$.

Let's consider the case whet $AC> AB$. In this case point $A'$ is positioned between $D$ and $C$. Denote midpoint of $DC$ as $E$. Then triangles $ACD$ and $B'CE$ are similar. Then $\angle B'EC=90°$. Triangles $B'EC$ and $B'ED$ are equal. Then $\angle B'DA'=\angle B'DE=\angle B'CE=\angle BCA$.

Ivan Kaznacheyeu
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