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