AD=5 AB=8 LK=√3 ise AL=?
m(DLA)=m(LAB)→|DL|=|DA|=5
|LC|=3
m(ABN)=m(LCN)→[AK]⊥[KC]
[DH]⊥[AK] olsun.
△DHK≡△LKC
|HL||DL|=|LK||LC|
x25=√33
x=10√33