Euler's Line of a medial triangle

I have the following problem with a comment below on the steps that I took so far.
Here is the example: Let triangle ABC be any triangle. The midpoints of the sides in Triangle ABC are labeled $A', B', C'$ of sides: $BC, CA, AB$. Let $D, K, I$ be the circumcenter, centroid and orthocenter, $D, K, I$. Let $D', K', I'$ be circumcenter, orthocenter of triangle $A'B'C'$.
$1.$Prove that Euler line of triangle ABC coincides with the Euler line of triangle $A'B'C'$
$2.$ Calculate $D'D/I'I$.
We know that $A′B′C′$ and $ABC$ is in a $1:2$ ratio since it is formed by the midpoints of the larger triangle called a medial triangle. The orthocenter of Triangle $ABC$ is also going to the same as the circumcenter of the $A′B′C′$.I can safely assume that the second part will be equal to 1. And I could probably use a dilation of $-1/2$ to prove that the medial triangle is the same as the larger one. Please add comments and suggestions.

• One person, many, many questions. Please, if you're having this much trouble with geometry, see the instructor. May 14, 2012 at 4:55
• What does -1 mean(on the left hand side next to the grey star and vertical arrows)? May 15, 2012 at 1:43
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• It's a "down-vote". The objective reason why this might happen is another user views this question as "unclear" or "not showing research effort". Alas, there are many subjective reasons that you might receive a downvote. I don't know when the vote happened, but if it was before you added your work, that could be the reason. May 15, 2012 at 2:05
• user31284, you can (should) edit your comments (especially those containing your work/efforts) into the body of the question so that those viewing will see the important information all together. May 15, 2012 at 2:13

The triangles $ABC$ and $A'B'C'$ have the same center of gravity (intersection of medians). Furthermore, you can prove that the circumcenter of $ABC$ is the orthocenter of $A'B'C'$ (draw a precise figure when in doubt). Therefore, the Euler lines of the two triangles are the same.
We can obtain $A'B'C'$ by performing an homothety of center $K$, the centroid, and ratio $-1/2$. Therefore, $DD' = 3/2KD$. Moreover $II' = 3/2 IK$. Now, in order to find $DD'/II'$ you just need $KD/IK$ and this comes from the ratios made by the points $I,K,D$ on the Euler line.
(Using notations like $I,K,D$ is really confusing. One generally notes $G$ for the center of gravity, $O$ the circumcenter and $H$ the orthocenter)