DB/BC=3/4=> DB/3=BC/4=k=> BD=3k si BC=4k
=>CD=k
Fie DM paralel cu AB, M apartine AC si DM perpendicular pe AC.
▲ABC, AB||DM=> conform teoremei fundamentale a asemanarii, ▲ABC eate asemenea cu ▲CMD=> MD/AB=CD/BC=MC/CA
Luam primele 2 proportii=>
MD/15=k/4k=>=>MD/15=1/4=>MD=15/4 (cm)
m(DAB)=m(DAC)=m(A)/2=90°/2=45°
In ▲ DMA, m(M)=90, m(MAD)=45°=> m(MDA)=45° (triungiul este drept. si isoscel)
=>MD=MA=15/4
CM/AC=MD/AB=>(AC-CM)/AC=(15-15/4)/AB=> AM/AC=45/4 • 1/15=> (15/4)/AC=3/4=> AC=5 (cm)
In ▲ABC, AB perpendicular pe BC,=> BC^2=AB^2+AC^2=>BC^2=225+25=>BC^2=250=>BC=5√10( cm)
P=15+5+5√10=20+5√10=5(4+√10) cm.
Fie AM perpendicular pe BC, M apartine BC
▲ABC, AB perpendicular pe AC si AM perpendicular pe BC, conform teoremei inaltimii rezulta:
AM=AB×AC/BC=>AM=(15×5)/5√10=>AM=15/√10=15√10/10=(3√10)/2.
Sper c-am fost de ajutor!
