As shown in the figure, in the quadrilateral ABCD, the point E is on BC, ∠ a + ∠ ade = 180 °, ∠ B = 78 ° and ∠ C = 60 ° to find the degree of ∠ EDC

As shown in the figure, in the quadrilateral ABCD, the point E is on BC, ∠ a + ∠ ade = 180 °, ∠ B = 78 ° and ∠ C = 60 ° to find the degree of ∠ EDC

It is proved that ∵ a + ∵ ade = 180 °, ∥ ab ∥ De, ∵ CED = ∵ B = 78 ° and ∵ C = 60 ° and ∵ EDC = 180 ° - (∵ CED + ∥ C) = 180 ° - (78 ° + 60 °) = 42 °

In trapezoidal ABCD, ad ∥ BC (AD ﹤ BC) AC and BD intersect at point O. if s △ OBC = 9 / 25s trapezoidal ABCD, calculate the perimeter ratio of △ oad to △ BOC

2:3
S triangle AOD × s triangle BOC = the square of s triangle AOB
(Note: use s ⊿ = (?) 189; absin θ)
And because s triangle AOB = s triangle doc
First, the BOC of s triangle is 4 / 25s, and then the square of perimeter ratio is the area ratio

In ladder ABCD, AD / / BC, diagonal AC, BD intersect at O, s △ abd = 4, s △ DBC = 10, find s △ AOD

16/3

[1] prove that the quadrilateral ABCD is a parallelogram, [2] if the area of △ AOB is 15cm & # 178;, calculate the area of quadrilateral ABCD

From the central symmetry, we can know that: OA = OC, OB = OD, angle AOB = angle cod, we can know that △ AOB is all equal to △ cod. So the angle ABO = angle CDO, so there is a straight line AB parallel to the straight line CD (the inner stagger angle is equal, two straight lines are parallel); similarly, we can prove that △ AOD is all equal to △ BOC, so there is an angle ADO = angle CBO, so ad parallel to BC; from ab parallel