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 ΔPQR, 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⟹UT=SV
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Let J=HF∩AD and L=CE∩KG
We have ∠AHJ=∠FDJ=∠BDE=∠ECA⟹PJ∥BL
Similarly, we also have BJ∥PL⟹BJPL is a parallelogram
Construct M=JL∩BP then M is the midpoint of JL
This means we are done if we can show that DE∥JL
Utilizing the Lemma, we get DF=EG⟹ΔDFJ=ΔELG⟹FJ∥=EL⟹EFJL is a parallelogram thus DE∥JL
Hence done