Teorema bisectoarei
Marian Tache
Liceul Teoretic W. ShakespeareTimisoara
March 29, 2015
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 1 / 11
Teorema bisectoarei
Theorem
Bisectoarea unui unghi al unuitriunghi determina pe laturaopusa segmente proportionalecu lungimile laturilor ceformeaza unghiul.
(AD − bisectoarea∠A⇔
BDDC = AB
AC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 2 / 11
Teorema bisectoarei
Theorem
Bisectoarea unui unghi al unuitriunghi determina pe laturaopusa segmente proportionalecu lungimile laturilor ceformeaza unghiul.(AD − bisectoarea∠A⇔
BDDC = AB
AC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 2 / 11
Teorema bisectoarei
Theorem
Bisectoarea unui unghi al unui triunghidetermina pe latura opusa segmenteproportionale cu lungimile laturilor ceformeaza unghiul.(AD − bisectoarea∠A⇔ BD
DC = ABAC
Proof.
CM⊥AD, MN‖BC ⇒⇒ [AM] ≡ [AC ], [MN] ≡ [DC ]4ABD ∼ 4AMN
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 3 / 11
Teorema bisectoarei
Theorem
Bisectoarea unui unghi al unui triunghidetermina pe latura opusa segmenteproportionale cu lungimile laturilor ceformeaza unghiul.(AD − bisectoarea∠A⇔ BD
DC = ABAC
Proof.
CM⊥AD, MN‖BC ⇒
⇒ [AM] ≡ [AC ], [MN] ≡ [DC ]4ABD ∼ 4AMN
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 3 / 11
Teorema bisectoarei
Theorem
Bisectoarea unui unghi al unui triunghidetermina pe latura opusa segmenteproportionale cu lungimile laturilor ceformeaza unghiul.(AD − bisectoarea∠A⇔ BD
DC = ABAC
Proof.
CM⊥AD, MN‖BC ⇒⇒ [AM] ≡ [AC ], [MN] ≡ [DC ]
4ABD ∼ 4AMN
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 3 / 11
Teorema bisectoarei
Theorem
Bisectoarea unui unghi al unui triunghidetermina pe latura opusa segmenteproportionale cu lungimile laturilor ceformeaza unghiul.(AD − bisectoarea∠A⇔ BD
DC = ABAC
Proof.
CM⊥AD, MN‖BC ⇒⇒ [AM] ≡ [AC ], [MN] ≡ [DC ]4ABD ∼ 4AMN
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 3 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c
Proof.
Notez BDDC = k Aplic teorema “k”:
AD = 11+kAB + k
1+kAC ⇔ AD = 11+BD
DC
AB +BDDC
1+BDDC
AC ⇔
⇔ AD = 11+AB
AC
AB +ABAC
1+ABAC
AC ⇔
⇔ AD = ACAB+ACAB + AB
AB+ACAC ⇔⇔ AD = b
b+cAB + cb+cAC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 4 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c
Proof.
Notez BDDC = k Aplic teorema “k”:
AD = 11+kAB + k
1+kAC ⇔ AD = 11+BD
DC
AB +BDDC
1+BDDC
AC ⇔
⇔ AD = 11+AB
AC
AB +ABAC
1+ABAC
AC ⇔
⇔ AD = ACAB+ACAB + AB
AB+ACAC ⇔⇔ AD = b
b+cAB + cb+cAC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 4 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c
Proof.
Notez BDDC = k Aplic teorema “k”:
AD = 11+kAB + k
1+kAC ⇔ AD = 11+BD
DC
AB +BDDC
1+BDDC
AC ⇔
⇔ AD = 11+AB
AC
AB +ABAC
1+ABAC
AC ⇔
⇔ AD = ACAB+ACAB + AB
AB+ACAC ⇔⇔ AD = b
b+cAB + cb+cAC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 4 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c
Proof.
Notez BDDC = k Aplic teorema “k”:
AD = 11+kAB + k
1+kAC ⇔ AD = 11+BD
DC
AB +BDDC
1+BDDC
AC ⇔
⇔ AD = 11+AB
AC
AB +ABAC
1+ABAC
AC ⇔
⇔ AD = ACAB+ACAB + AB
AB+ACAC ⇔⇔ AD = b
b+cAB + cb+cAC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 4 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c
Proof.
Notez BDDC = k Aplic teorema “k”:
AD = 11+kAB + k
1+kAC ⇔ AD = 11+BD
DC
AB +BDDC
1+BDDC
AC ⇔
⇔ AD = 11+AB
AC
AB +ABAC
1+ABAC
AC ⇔
⇔ AD = ACAB+ACAB + AB
AB+ACAC ⇔
⇔ AD = bb+cAB + c
b+cAC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 4 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c
Proof.
Notez BDDC = k Aplic teorema “k”:
AD = 11+kAB + k
1+kAC ⇔ AD = 11+BD
DC
AB +BDDC
1+BDDC
AC ⇔
⇔ AD = 11+AB
AC
AB +ABAC
1+ABAC
AC ⇔
⇔ AD = ACAB+ACAB + AB
AB+ACAC ⇔⇔ AD = b
b+cAB + cb+cAC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 4 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c
(BE − bisectoarea∠B ⇔BE = cBC+aBA
c+a(CF − bisectoarea∠C ⇔CF = aCA+BCB
a+b
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 5 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c(BE − bisectoarea∠B ⇔BE = cBC+aBA
c+a
(CF − bisectoarea∠C ⇔CF = aCA+BCB
a+b
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 5 / 11
Teorema bisectoarei
Theorem
(AD − bisectoarea∠A⇔AD = bAB+cAC
b+c(BE − bisectoarea∠B ⇔BE = cBC+aBA
c+a(CF − bisectoarea∠C ⇔CF = aCA+BCB
a+b
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 5 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c .
Proof.
AI = αAD; BI = βBE ;
AD = bAB+cACb+c ; BE = cBC+aBA
c+a ; folosescAB = AI − BI
AB = αbb+cAB + αc
b+cAC −βcc+aBC −
βac+aBA; folosescAC = AB − BC
AB = αbb+cAB + αc
b+cAB + αcb+cBC −
βcc+aBC −
βac+aBA
AB(α + βa
c+a − 1)
+ BC(
αcb+c −
βcc+a
)= 0
α = b+ca+b+c ⇒ AI = bAB+cAC
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 6 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c .
Proof.
AI = αAD; BI = βBE ;
AD = bAB+cACb+c ; BE = cBC+aBA
c+a ; folosescAB = AI − BI
AB = αbb+cAB + αc
b+cAC −βcc+aBC −
βac+aBA; folosescAC = AB − BC
AB = αbb+cAB + αc
b+cAB + αcb+cBC −
βcc+aBC −
βac+aBA
AB(α + βa
c+a − 1)
+ BC(
αcb+c −
βcc+a
)= 0
α = b+ca+b+c ⇒ AI = bAB+cAC
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 6 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c .
Proof.
AI = αAD; BI = βBE ;
AD = bAB+cACb+c ; BE = cBC+aBA
c+a ; folosescAB = AI − BI
AB = αbb+cAB + αc
b+cAC −βcc+aBC −
βac+aBA; folosescAC = AB − BC
AB = αbb+cAB + αc
b+cAB + αcb+cBC −
βcc+aBC −
βac+aBA
AB(α + βa
c+a − 1)
+ BC(
αcb+c −
βcc+a
)= 0
α = b+ca+b+c ⇒ AI = bAB+cAC
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 6 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c .
Proof.
AI = αAD; BI = βBE ;
AD = bAB+cACb+c ; BE = cBC+aBA
c+a ; folosescAB = AI − BI
AB = αbb+cAB + αc
b+cAC −βcc+aBC −
βac+aBA; folosescAC = AB − BC
AB = αbb+cAB + αc
b+cAB + αcb+cBC −
βcc+aBC −
βac+aBA
AB(α + βa
c+a − 1)
+ BC(
αcb+c −
βcc+a
)= 0
α = b+ca+b+c ⇒ AI = bAB+cAC
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 6 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c .
Proof.
AI = αAD; BI = βBE ;
AD = bAB+cACb+c ; BE = cBC+aBA
c+a ; folosescAB = AI − BI
AB = αbb+cAB + αc
b+cAC −βcc+aBC −
βac+aBA; folosescAC = AB − BC
AB = αbb+cAB + αc
b+cAB + αcb+cBC −
βcc+aBC −
βac+aBA
AB(α + βa
c+a − 1)
+ BC(
αcb+c −
βcc+a
)= 0
α = b+ca+b+c ⇒ AI = bAB+cAC
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 6 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c .
Proof.
AI = αAD; BI = βBE ;
AD = bAB+cACb+c ; BE = cBC+aBA
c+a ; folosescAB = AI − BI
AB = αbb+cAB + αc
b+cAC −βcc+aBC −
βac+aBA; folosescAC = AB − BC
AB = αbb+cAB + αc
b+cAB + αcb+cBC −
βcc+aBC −
βac+aBA
AB(α + βa
c+a − 1)
+ BC(
αcb+c −
βcc+a
)= 0
α = b+ca+b+c ⇒ AI = bAB+cAC
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 6 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c .
Proof.
AI = αAD; BI = βBE ;
AD = bAB+cACb+c ; BE = cBC+aBA
c+a ; folosescAB = AI − BI
AB = αbb+cAB + αc
b+cAC −βcc+aBC −
βac+aBA; folosescAC = AB − BC
AB = αbb+cAB + αc
b+cAB + αcb+cBC −
βcc+aBC −
βac+aBA
AB(α + βa
c+a − 1)
+ BC(
αcb+c −
βcc+a
)= 0
α = b+ca+b+c ⇒ AI = bAB+cAC
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 6 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c .
I -intersectia bisectoarelor4ABC ⇒ BI = cBC+aBA
a+b+c . I -intersectia bisectoarelor4ABC ⇒ CI = aCA+bCB
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 7 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c . I -intersectia bisectoarelor4ABC ⇒ BI = cBC+aBA
a+b+c .
I -intersectia bisectoarelor4ABC ⇒ CI = aCA+bCB
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 7 / 11
Teorema 1
Theorem
I - intersectia bisectoarelor4ABC ⇒ AI = bAB+cAC
a+b+c . I -intersectia bisectoarelor4ABC ⇒ BI = cBC+aBA
a+b+c . I -intersectia bisectoarelor4ABC ⇒ CI = aCA+bCB
a+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 7 / 11
Vectorul de pozitie al centrului cercului ınscris
Theorem
I - intersectia bisectoarelor4ABC ⇒ rI = arA+brB+crC
a+b+c .
Proof.
AI = bAB+cACa+b+c
rI − rA = brB−brA+crC−crAa+b+c
rI = brB−brA+crC−crA+arA+brA+crAa+b+c
rI = arA+brB+crCa+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 8 / 11
Vectorul de pozitie al centrului cercului ınscris
Theorem
I - intersectia bisectoarelor4ABC ⇒ rI = arA+brB+crC
a+b+c .
Proof.
AI = bAB+cACa+b+c
rI − rA = brB−brA+crC−crAa+b+c
rI = brB−brA+crC−crA+arA+brA+crAa+b+c
rI = arA+brB+crCa+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 8 / 11
Vectorul de pozitie al centrului cercului ınscris
Theorem
I - intersectia bisectoarelor4ABC ⇒ rI = arA+brB+crC
a+b+c .
Proof.
AI = bAB+cACa+b+c
rI − rA = brB−brA+crC−crAa+b+c
rI = brB−brA+crC−crA+arA+brA+crAa+b+c
rI = arA+brB+crCa+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 8 / 11
Vectorul de pozitie al centrului cercului ınscris
Theorem
I - intersectia bisectoarelor4ABC ⇒ rI = arA+brB+crC
a+b+c .
Proof.
AI = bAB+cACa+b+c
rI − rA = brB−brA+crC−crAa+b+c
rI = brB−brA+crC−crA+arA+brA+crAa+b+c
rI = arA+brB+crCa+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 8 / 11
Vectorul de pozitie al centrului cercului ınscris
Theorem
I - intersectia bisectoarelor4ABC ⇒ rI = arA+brB+crC
a+b+c .
Proof.
AI = bAB+cACa+b+c
rI − rA = brB−brA+crC−crAa+b+c
rI = brB−brA+crC−crA+arA+brA+crAa+b+c
rI = arA+brB+crCa+b+c .
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 8 / 11
Consecinte
Intr-un triunghibisectoarele suntconcurente.
aAI + bBI + cCI = 0
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 9 / 11
Consecinte
Intr-un triunghibisectoarele suntconcurente.
aAI + bBI + cCI = 0
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 9 / 11
Aplicatia 1
In triunghiul4ABC , (AD), (BE ), (CF )sunt bisectoarele unghiurilor.Aratati ca dacaAD + BE + CF = 0, atunci4ABC este echilateral.
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 10 / 11
Teorema bisectoarei exterioare
Theorem
(BD −bisectoarea exterioar a a∠B ⇔DADC = AB
BC
Proof.
Construim AE ||BC ,E ∈ BD,
⇒ ∠AEB ≡ ∠DBx(alt.int) s i ∠DBA ≡ ∠DBx⇒ ∠AEB ≡ ∠DBx ⇒4AEB isos.⇒ AE = ABCum AE ||BC ⇒4DEA ∼ 4DBC⇒ DA
DC = EABC = AB
BC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 11 / 11
Teorema bisectoarei exterioare
Theorem
(BD −bisectoarea exterioar a a∠B ⇔DADC = AB
BC
Proof.
Construim AE ||BC ,E ∈ BD,⇒ ∠AEB ≡ ∠DBx(alt.int) s i ∠DBA ≡ ∠DBx
⇒ ∠AEB ≡ ∠DBx ⇒4AEB isos.⇒ AE = ABCum AE ||BC ⇒4DEA ∼ 4DBC⇒ DA
DC = EABC = AB
BC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 11 / 11
Teorema bisectoarei exterioare
Theorem
(BD −bisectoarea exterioar a a∠B ⇔DADC = AB
BC
Proof.
Construim AE ||BC ,E ∈ BD,⇒ ∠AEB ≡ ∠DBx(alt.int) s i ∠DBA ≡ ∠DBx⇒ ∠AEB ≡ ∠DBx ⇒4AEB isos.⇒ AE = AB
Cum AE ||BC ⇒4DEA ∼ 4DBC⇒ DA
DC = EABC = AB
BC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 11 / 11
Teorema bisectoarei exterioare
Theorem
(BD −bisectoarea exterioar a a∠B ⇔DADC = AB
BC
Proof.
Construim AE ||BC ,E ∈ BD,⇒ ∠AEB ≡ ∠DBx(alt.int) s i ∠DBA ≡ ∠DBx⇒ ∠AEB ≡ ∠DBx ⇒4AEB isos.⇒ AE = ABCum AE ||BC ⇒4DEA ∼ 4DBC
⇒ DADC = EA
BC = ABBC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 11 / 11
Teorema bisectoarei exterioare
Theorem
(BD −bisectoarea exterioar a a∠B ⇔DADC = AB
BC
Proof.
Construim AE ||BC ,E ∈ BD,⇒ ∠AEB ≡ ∠DBx(alt.int) s i ∠DBA ≡ ∠DBx⇒ ∠AEB ≡ ∠DBx ⇒4AEB isos.⇒ AE = ABCum AE ||BC ⇒4DEA ∼ 4DBC⇒ DA
DC = EABC = AB
BC
Marian Tache (Liceul Teoretic W. Shakespeare Timisoara)Teorema bisectoarei March 29, 2015 11 / 11
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