27.03.2012 - geometrie [cls VII-VIII]

mircea.lascu
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27.03.2012 - geometrie [cls VII-VIII]

Mesaj 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 $PT = QS$.
Nelu Cararus
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Re: 27.03.2012 - geometrie [cls VII-VIII]

Mesaj 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 $PT =QS$
Virgil Nicula
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Re: 27.03.2012 - geometrie [cls VII-VIII]

Mesaj 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 $XQ=YP$ $\Longrightarrow$ $\triangle PYT\equiv\triangle QXS$ $\Longrightarrow$ $PT=QS$ .
Ultima oară modificat Mie Iul 09, 2014 3:22 pm de către Virgil Nicula, modificat de 2 ori în total.
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sunken rock
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Re: 27.03.2012 - geometrie [cls VII-VIII]

Mesaj 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
A blind man sees the details better.
mihai miculita
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Re: 27.03.2012 - geometrie [cls VII-VIII]

Mesaj 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
".
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