It is known that the perimeter of the parallelogram ABCD is 50cm, the diagonals intersect at point O, the perimeter of △ AOB is 5cm longer than that of △ BOC, and the lengths of AB and BC are calculated No picture

It is known that the perimeter of the parallelogram ABCD is 50cm, the diagonals intersect at point O, the perimeter of △ AOB is 5cm longer than that of △ BOC, and the lengths of AB and BC are calculated No picture


AB + BC = 25 copies, △ AOB perimeter - △ BOC perimeter = 5, ab-bc = 5
Ab-bc = 5
So if we add two to one, we get 2Ab = 30, then we get AB = 15, BC = 10



In the parallelogram ABCD, De is perpendicular to AB and E, BF is perpendicular to AD and F
1. Prove that AD / de = AB / BF
2. The perimeter of parallelogram ABCD is 12, ad: de = 5:2. Calculate the value of de + BF


(1) rt ⊿ ade ⊿ RT ⊿ AFB, AD / AB = de / BF, that is, AD / de = AB / BF
(2) let de = 2x. Then ad = 5x. AB = 6-5x
∵AD/AB=DE/BF.∴DE/(DE+BF)=AD/(AD)+AB).
DE+BF=(AB+AD)DE/AD=[(6-5x)+5x]×2x/5x=12/5=2.4