Question

In triunghiul ABC oarecare AB=21 cm, BC=24 cm si AC=15 cm , [AD bisectoarea BAC, iar DE||AB. Aflaţi:
a) Lungimea segmentului [BD] şi [CD];
b) Perimetrul triunghiului CDE.

Answer 1

Răspuns:

a)

BD = 14 cm

CD = 10 cm

b)

P(ΔCDE) = 25 cm

Explicație pas cu pas:

A) Aplicam Teorema Bisectoarei care spune cam asa :

BD/DC = AB/AC <=>

BD/(DC+BD) = AB/(AC+AB) <=>

BD/BC = AB/(AC+AB)

Inlocuim valorile segmentelor

BD/24 = 21/36

BD = 21×24/36

BD = 504/36

BD = 14 cm

CD = BC - BD = 24cm - 14cm

CD = 10 cm

B)

DE || AB => conform Teoremei lui Thales

CD/BC = EC/AC

10/24 = EC/15

=> EC = 10×15/24 = 150/24 => EC =  6,25 cm

ED/AB = CD/AB

ED/21 = 10/24

=> ED = 10×21/24 = 210/24 => ED = 8,75 cm

P(ΔCDE) = CD+CE+ED

= 10cm + 6,25cm + 8,75cm

= 25 cm

Answer 2

$\it Teorema\ bisectoarei \Rightarrow \dfrac{BD}{CD}=\dfrac{AB}{AC} \Rightarrow \dfrac{BD}{CD}=\dfrac{\ 21^{(3}}{15} \Rightarrow\dfrac{BD}{CD}=\dfrac{7}{5} \Rightarrow\\ \\ \\ \Rightarrow\dfrac{BD}{CD}=1,4 \Rightarrow BD=1,4CD\ \ \ \ \ (1) \\ \\ \\ Dar,\ BD+CD=BC \Rightarrow BD+CD=24\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow 1,4CD+CD+24 \Rightarrow 2,4CD=24 \Rightarrow CD=10\ cm$

$\it BD=BC-CD=24-10=14\ cm\\ \\ \\ b)\ DE||AB \stackrel{T.f.a.}{\Longrightarrow}\ \Delta ABC\sim\Delta EDC \Rightarrow \dfrac{\mathcal{P}_{ABC}}{\mathcal{P}_{EDC}}=k\ (raportul\ de\ asem\breve{a}nare )\ \ \ (*)\\ \\ \\ \Delta ABC\sim\Delta EDC \Rightarrow k=\dfrac{BC}{CD}=\dfrac{24}{10}=2,4$

Acum, relația (*) devine:

$\it \dfrac{21+24+15}{\mathcal{P}_{EDC}}=2,4 \Rightarrow \dfrac{60}{\mathcal{P}_{EDC}}=2,4 \Rightarrow \mathcal{P}_{EDC}=\dfrac{^{5)}60}{2,4}=\dfrac{300}{12}=25\ cm$