Răspuns:
Explicație pas cu pas:
ΔABC, ∡B=90°, AB=2cm, m(∡ACB)=30°, M∈[BC], CM=2(√3-1)cm.
d(B,AC)=??? Trasam BD⊥AC, D∈AC. ⇒BD=d(B,AC). In ΔABC, m(∡ACB)=30°, ⇒m(∡CAB)=60°, atunci in ΔABD, dreptunghic in D, m(∡ABD)=30°, ⇒AD=(1/2)·AB=(1/2)·2=1cm. Dupa T.Pitagora, ⇒BD²=AB²-AD²=2²-1²=3. Atunci BD=√3cm=d(B,AC).
d(M,AC)=??? Trasam ME⊥AC, E∈AC. Atunci d(M,AC)=ME. In ΔCME, m(∡ECM)=30°, deci ME=(1/2)·CM=(1/2)·2(√3-1)cm=(√3-1)cm=d(M,AC).
d(C,AM)=??? Trasam dreapta AM. Trasam CN⊥AM, N∈AM. Atunci CN=d(C,AM). ΔCNM≅ΔABM, dreptunghice cu cate un unghi ascutit congruente (opuse la varf). Atunci CN/AB=CM/AM=NM/BM.
In ΔABC, AB=2cm, ⇒AC=2·AB=2·2=4cm. T.P. ⇒BC²=AC²-AB²=4²-2²=16-4=12=4·3. Deci BC=√(4·3)=2√3cm. Atunci BM=BC-CM=2√3-2(√3-1)=2√3-2√3+2=2. Deci AB=2cm=BM, ⇒ΔABM dreptunghic isoscel. Atunci si ΔCNM este dreptunghic isoscel, deci NM=CN
Din ΔABM, AM=2√2
Din CN/AB=CM/AM, ⇒CN/2=2(√3-1)/(2√2)=(√3-1)/√2=√2(√3-1)/ 2.
Atunci CN=√2(√3-1)/ 2cm=d(C,AM).
d(B,AM)=??? Trasam BP⊥AM, P∈AM, atunci d(B,AM)=BP.
ΔCNM≅ΔBPM, dreptunghice cu cate un unghi ascutit congruente (opuse la varf). Atunci CN/BP=CM/BM. ⇒
[tex]\frac{\sqrt{2}(\sqrt{3}-1)}{2} :BP=\frac{2(\sqrt{3}-1) }{2} ,~BP=\frac{\sqrt{2}(\sqrt{3}-1)}{2}:\frac{2(\sqrt{3}-1) }{2}=\frac{\sqrt{2}(\sqrt{3}-1)}{2}*\frac{1}{\sqrt{3}-1 }=\frac{\sqrt{2} }{2} cm[/tex]
Deci d(B,AM)=√2/2cm
