teorema bisectoarei.pdf

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