Let H be the orthocenter of the sharp triangle ABC

Question

Let $H$ be the orthocenter of acute triagle $ABC$. On the single direction $(HA$, $(HB$ and $(HC$ is considered points $D$, $E$ and $F$ such that: $HD = BC$, $HE = AC$ and $HF = AB$.

Let it be shown that:

1. if $M$, $N$ and $P$ are the midpoint$[BC]$, $[CA]$ and $[AB]$, then: $DE = 2CP$, $EF = 2AM$ and $DF = 2BN$.
2. $H$ is the center of gravity of the triangle $DEF$.

Answer 1

You can prove first part with the law of cosine. Let $BC = a$, $CA = b$ and $AB = c$. If $\angle ACB = x$, then we have (observe triangle $DEH$):

\begin{align} ED^2 & = a^2+b^2-2ab\cos (180-x) \\ & = a^2+b^2+2ab \cos x\\ & = a^2+b^2 +(a^2+b^2-c^2)\\ & = 2a^2+2b^2-c^2 \end{align}

On the other hand we have (observe triagle $BCP$) \begin{eqnarray} CP^2 & = & a^2+{c^2\over 4}-2a{c\over 2}\color{\red}{\cos (\beta) }\\ & = & a^2+{c^2\over 4}-ac \color{\red}{a^2+c^2-b^2\over 2ac}\\ & = &...\\ & = & {2a^2+2b^2-c^2\over 4}\\ & = & {ED^2\over 4} \end{eqnarray} For red relation observe the triangle $ABC$

Answer 2

Main hint: Construct point T such that $CBTA$ is a parallelogram. Then, $2CP = CT$.

Hint to 1: Angle chase + Side chase triangle $HDE$.
Show that this is congruent to another triangle involving $CT$.
Hence conclude that $ DE = CT = 2CP$.

$\triangle HDE \sim \triangle BTC$ so $DE = TC = 2CP$.

Hint to 2: Let midpoint of $DE$ be $X$. Angle chase + Side chase triangle $EHX$.
Show that this triangle is congruent to another triangle involving $CP$.
Hence conclude that $ EHX$ is a straight line.

$\triangle EHX \sim \triangle CBP $. Actually, this is just the previous triangle, using the midpoint on $DE$ and $CT$. Hence, $\angle EHX = \angle BAC $, so $EHX$ is a straight line.

Corollary: Since $EX$ is the median, so $H$ is the centroid of $DEF$.

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Diagram not drawn to scale because I was too lazy to ensure $HD=BC$.