In any triangle ABC, D, E and F are the midpoints of AB, BC and AC and from which perpendiculars are dropped on sides AB, BC and AC.The area of triangle ABC is S. Find the area of the polygon DHEIFG in terms of S.
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First, note that triangle $ADF, DBE, FEC, EDF$ are all congruent. SO $S_{DEF}={1\over4}S$.
Second, note that $H$ is the orthocenter of $BDE$, $I$ is the orthocenter of $FEC$ and $G$ is the orthocenter of $ADF$, and also by the previously stated congruence, we have triangles $GDF,DHE,FEI$ made up exactly one copy of $ADF$ which has sum ${1\over4}S$.
Hence totally the hexagon has area ${1\over2}S$.
cr001
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@croo1 Many thanks! – pirsquare Oct 26 '15 at 09:50
