In the acute-angled triangle $ABC$, let $CD, AE$ be the altitudes. Points $F$ and $G$ are the projections of $A$ and $C$ on the line $DE$, respectively, $H$ and $K$ are the projections of $D$ and $E$ on the line $AC$, respectively. The lines $HF$ and $KG$ intersect at point $P$. Prove that line $BP$ bisects the segment $DE$.
Problem
Source:
Tags: geometry, projections, bisects segment, Kharkiv, altitudes
BestChoice123
22.03.2020 17:04
Could you check your problem again? I think something has gone wrong...
parmenides51
22.03.2020 17:11
the altitudes are $CD, AE$ and not $CD,BE$, thanks
BestChoice123
23.03.2020 05:19
: In $\Delta PQR$, let $QS,RT$ be the altitudes. Let $O$ be the midpoint of $QR$. Points $U,V,W$ are the projections of $Q,R,O$ onto line $TS$ then $TU=SV$
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Proof
Note that $QUVR$ is a right trapezoid thus $W$ is the midpoint of $UV$
Since $QTSR$ is a cyclic quadrilateral with circumcenter $O$, we have $WT=WS\implies UT=SV$
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Let $J=HF\cap AD$ and $L=CE\cap KG$
We have $\angle AHJ=\angle FDJ=\angle BDE=\angle ECA\implies PJ\parallel BL$
Similarly, we also have $BJ\parallel PL\implies BJPL$ is a parallelogram
Construct $M=JL\cap BP$ then $M$ is the midpoint of $JL$
This means we are done if we can show that $DE\parallel JL$
Utilizing the Lemma, we get $DF=EG\implies \Delta DFJ=\Delta ELG\implies FJ\parallel =EL\implies EFJL$ is a parallelogram thus $DE\parallel JL$
Hence done