AC is the diagonal of parallelogram ABCD. Prove AB = CD BC = da

AC is the diagonal of parallelogram ABCD. Prove AB = CD BC = da

∵ AD / / BC AB / / CD ≌ DAC = BCA ≌ BCA ≌ AB = CD BC = Da

If ad = 6, ab = 10 and diagonal DB ⊥ ad in quadrilateral ABCD, the area of parallelogram ABCD is____

Ad ⊥ BD, according to Pythagorean theorem, BD = 8,
S△ABD=AD*BD/2=6*8/2=24,
S parallelogram = 2S △ abd = 48

In the parallelogram ABCD, ad = 3 √ 3, OB = 3, diagonal AC = 12
O is the intersection of the diagonals

O should be the midpoint of the diagonal
According to the conditions
AD=3√3,
AC=12 => OA = 6
OB=3 => OD = 3
It can be found that,
AD*AD = 27
OA*OA = 36
Od * od = 9 conforms to Pythagorean theorem
Ad is perpendicular to BD
Sabcd= 2Sabd = 2 x 1/2 x AD x BD = 3√3 x 6 = 18√3