The quadrilateral ABCD is a diamond, the edge AB is on the y-axis, the vertex D is in the first quadrant, B (0, - 4), C (3,0) are known (1) Finding the coordinate of point a (2) Finding the inverse proportion function relation of passing through point d

The quadrilateral ABCD is a diamond, the edge AB is on the y-axis, the vertex D is in the first quadrant, B (0, - 4), C (3,0) are known (1) Finding the coordinate of point a (2) Finding the inverse proportion function relation of passing through point d

（1）∵B(0,-4),C(3,0)
∴BC＝√（4²+3²）＝5
The quadrilateral ABCD is a diamond,
∴AB＝BC＝5
The coordinate of point a is (0,1)
（2）∵C(3,0),CD＝5
∴D（3,5）
Let the inverse scale function passing through point d be y = K / x, and D (3,5) be substituted into,
K / 3 = 5
∴k=15
The inverse proportional function through point D is y = 15 / X

In diamond ABCD, AC = 6cm, BD = 8cm. Find the distance between parallel line AB and CD

As shown in the figure, in the diamond ABCD
In the triangle abd, let AC and BD intersect at o
Make a vertical line to CD through point a. the vertical foot is p
In ACD, the triangle area is 6 * 4 * 0.5 = 12
If co = 3, OD = 4, the side length of diamond is 5, then the triangle area is equal, 12 = 5 * 0.5 * AP
Then, AP can be calculated as = 4.8
So the distance between AB and CD is 4.8

As shown in the figure, the square ABCD is known, and the quadrilateral aefc is a diamond, EH is perpendicular to AC and at point h, and eh = 1 / 2FC is proved
Don't use trigonometric functions

∵ quadrilateral aefc is rhombic
∴AC∥EF,FC=AC
A quadrilateral ABCD is a square
∴OC=OA=OD=OB=1/2AC
BD⊥AC
∵EH⊥AC
∴EH∥BD,
∵AC∥BF
The obeh is a rectangle
∴EH=OB
∴FC=AC=2OB=2EH
∴FC=2EH
That is eh = 1 / 2FC

As shown in the figure, it is known that the quadrilateral ABCD is a square, the diagonal AC and BD intersect at O, the quadrilateral aefc is a diamond, eh ⊥ AC, and the perpendicular foot is h. verification: eh = 12fc

It is proved that in square ABCD, AC ⊥ BD, AC = BD, OB = 12bd = 12ac, and ∵ quadrilateral aefc is rhombus, ∵ AC = CF, AC ∥ EF, ∵ eh ⊥ AC, ∵ DBC = ∵ abd = ∵ CBF = 45 °, ∵ boh = ∵ ohe = ∵ OBE = 90 °, ∵ quadrilateral beho is rectangle, ∵ eh = ob, ∵ eh = 12ac = 12CF