d(A,BC)=h
d(E,AB)=h1
h1/h=AD/BC
BD=AB-AD=x
h1=h(AB-x)/BC=m-k.x
(k,m: hang so)
k=AB.h/BC; m=h/BC
m/k=AB
h1.x=S∆BDE
h1.x=x(m-kx)
S∆bdemax =>[x(m-kx) ] max
<=>[-kx^2+mx) ]max
=>x=m/(2k)=AB/2
=>D trung diem AB
kẻ EI vuông góc với AB
có Sbde=1/2.EI.BD
Sade=1/2.EI.AD
Sade/Sbde=AD/BD
Sade/Sabc=AD^2/AB^2
suy ra Sbde/Sabc = AD.BD/AB^2
biet tg ABC ko doi, AB ko doi
suy ra Sbde max khi AD.BD max khi AD=BD