Problem
Take a point $P$ in an equilateral triangle $ABC$.
Let $D, \ E$ and $F$ be the feet of perpendicular lines drawn from $P$ to sides $BC, \ AB$ and $CA$, respectively.
When $P$ is on the line segment joining the midpoints of $AB$ and $AC$, prove that $PD=PE+PF$.
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Solution
From the previous problem $0164$,
$$PD+PE+PF=AH,$$
$$∴ \ AH-PD=PE+PF.$$
However, since $PD=LH$,
$$AH-LH=PE+PF,$$
$$∴ \ AL=PE+PF.$$
Since $△ANL∼△ACH$ and $AN∶AC=1∶2$,
$$AL:AH=1:2,$$
$$∴ \ AL =LH,$$
$$∴ \ PD=AL,$$
$$∴ PD=PE+PF.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, pp.37-38.