Răspuns:
DC=(√6-√2)cm; BC=√6cm.
Explicație pas cu pas:
AD bisectoare, deci ∡CAD=30°. DN⊥AC, ⇒din ΔADN, DN=(1/2)·AD.
Fie DN=x, atunci AD=2x. Atunci AN²=AD²-DN²=(2x)²-x²=4x²-x²=3x². Deci AN=x√3. ΔAND≡ΔAMD, dupa ipotenuza comuna si un unghi ascutit, deci AN=AM=x√3, DN=DM=x. Atunci in ΔBMD, BM=DM=x, deoarece ∡B=∡BDM=45°. Atunci BD=x√2.
Dupa Teorema Bisectoarei, ⇒DC/DB=AC/AB
[tex]DC=\frac{DB*AC}{AB}=\frac{x\sqrt{2}*2 }{x*(\sqrt{3}+1)} =\frac{2\sqrt{2} }{\sqrt{3}+1 }=\frac{2\sqrt{2}(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{2\sqrt{2}(\sqrt{3}-1)}{(\sqrt{3})^{2}-1^{2} }=\frac{2\sqrt{2}(\sqrt{3}-1)}{3-1}=\sqrt{2}(\sqrt{3}-1) =\sqrt{6}-\sqrt{2}[/tex]
CN=CA-AN=2-x√3. Din ΔDNC, ⇒DN²+NC²=CD², ⇒x²+(2-x√3)²=(√6-√2)², ⇒x²+2²-2·2·x√3+(x√3)²=(√6)²-2·√6·√2+(√2)², ⇒x²+4-4x√3+3x²=6-2√12+2, ⇒4x²-4x√3+4-8+2√(4·3)=0, ⇒4x²-4x√3+4(√3-1)=0, |:4, ⇒x²-x√3+(√3-1)=0
Δ=(-√3)²-4·(√3-1)=3-4√3+4=7-4√3=2²-2·2·√3+(√3)²=(2-√3)²
De unde x=(√3+(2-√3))/2=1. Am verificat a doua valoare x=√3-1 nu-i valabila.
Deci DN=1=x. Atunci BD=x√2=1·√2=√2
Deci BC=BD+DC=√2+√6-√2=√6cm.
Supliment d(A,BC)=???
Trasam AE⊥BC, atunci ΔABE dreptunghic in E. Deoarece ∡B=45°, ⇒∡BEA=45°. Atunci AE=BE=y
Din ΔABE, AB=AM+BM=x√3+x√2=√3+√2. T.P. ⇒AE²+BE²=AB²
y²+y²=(√3+√2)², ⇒2y²=(√3+√2)², ⇒y²=(√3+√2)²/2
Deci y=(√3+√2)/√2=√2(√3+√2)/ 2=(√6+2)/2=d(A,BC)
