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27.03.2012 - geometrie [cls VII-VIII]

MesajScris: Lun Mar 26, 2012 8:51 am
de mircea.lascu
Fie \tr ABC ascutitunghic si H ortocentrul sau. Fie (AD si (HE bisectoarele unghiurilor \widehat{BAC} si \widehat{BHC} cu D,E\in (BC). Fie P si Q proiectiile lui E pe AB respectiv AC, iar S si T proiectiile lui D pe BH respectiv CH. Aratati ca PT = QS.

Re: 27.03.2012 - geometrie [cls VII-VIII]

MesajScris: Joi Iun 07, 2012 8:05 pm
de Nelu Cararus
Pentru inceput vom demonstra ca AD si EH sunt paralele:
Fie A_1,B_1,C_1 - picioarele inaltimilor din virfurile A,B, repectiv C. A_1,B_1,C_1\in (BC),(AC),(AB)
1) Este evident ca \angle BHC = 180 - \angle BAC,deoarece patrulaterul AC_1HB_1 - inscriptibil. Deci m(\angle EHC)= 90-\frac{1}{2}m(\angle BAC).
\angle HCA = \angle ABH (CB_1C_1C -inscriptibil):
m(\angle HCE) = m(\angle ACB) - m(\angle ABH) = 90 - m(\angle ABC) \Longrightarrow m(\angle HEC) = \frac{1}{2}m(\angle BAC) + m(\angle ABC).
Pe de alta parte m(\angle ADC) = 180 - \frac{1}{2}m(\angle BAC) - m(\angle ACB)= \frac{1}{2}m(\angle BAC) + m(\angle ABC) \Longrightarrow AD\parallel EH.
2) Fie K si L proiectiile punctului D pe AB si respectiv AC,
\angle BAC =\alpha , \angle ABC = \beta si \angle ACB = \gamma
3) Cum \angle KDA = \angle LDA , PE\parallel KD \parallel    C_1H , DL\parallel EQ \parallel B_1H , atunci
\angle PEH = \angle QEH si PC_1 = B_1Q , dar C_1T = KD ,
B_1S = LD , DK = DQ , atunci:
C_1T = B_1S. Din congruenta: \triangle PC_1T \equiv \triangle QB_1S rezulta PT =QS

Re: 27.03.2012 - geometrie [cls VII-VIII]

MesajScris: Mar Oct 02, 2012 8:39 am
de Virgil Nicula
PP. Let \tr ABC be an acute triangle with the orthocenter H . Let [AD , [HE be the bisectors of \widehat{BAC} , \widehat{BHC} respectively , where \{D,E\}\subset (BC) . Denote
the projections P , Q of the point E on AB , AC respectively and the projections S , T of the point D on BH , CH respectively. Prove that PT = QS .


Proof. Denote X\in BH\cap AC and Y\in CH\cap AB . Define the projections M , N of the point D on AB , AC respectively and the projections U , V of the point E on BH , CH respectively. Observe that DM=DN , EU=EV and DMYT , DNXS , EPYV , EQXU are rectangles. Therefore, YT=DM=DN=XS , YP=EV=EU=XQ . In conclusion, YT=XS and XQ=YP \Longrightarrow \triangle PYT\equiv\triangle QXS \Longrightarrow PT=QS .

Re: 27.03.2012 - geometrie [cls VII-VIII]

MesajScris: Dum Apr 14, 2013 4:18 pm
de sunken rock
Basically same as above, little bit shorter:

1) Well known AD\parallel HE.
2) Let X\in AC\cap BH, Y\in AB\cap CH. PY and QX are projections of HE onto AB, AC respectively; from 1) we infer PY\equiv QX \ (\ 1\ ).
3) SD\parallel AC and DT\parallel AB mean TY and SX are equal to the distances of D to AB, AC respectively, but AD is angle bisector, so TY\equiv SX\ (\ 2\ ).
4) (1)\wedge (2)\implies\Delta TYP\cong\Delta SXQ hence PT\equiv SQ.

Best regards,
sunken rock

Re: 27.03.2012 - geometrie [cls VII-VIII]

MesajScris: Sâm Apr 19, 2014 1:58 pm
de mihai miculita
O mica observatie:
Notand cu H_b si H_c piciarele inaltimilor duse din varfurile B si C in patrulaterul ABCD avem: \widehat{AH_bH}\equiv\widehat{AH_cH} (=90^0)..
Asa ca, paralelismul dreptelor: AD si HE (bisectoarea unghiului \widehat{H_bHH_c}), rezulta direct din teorema:

"Daca un patrulater are 2 unghiuri opuse congruente, atunci bisectoarele
celorlalte doua unghiuri opuse ale acestui patrulater sunt paralele
".