As shown in the figure, the vertex B of square ABCD is the bisector BF of diagonal AC, the point E is the point on BF, and the quadrilateral aefc is a diamond, eh ⊥ AC, and the perpendicular foot is h EH = 1 / 2cf,

As shown in the figure, the vertex B of square ABCD is the bisector BF of diagonal AC, the point E is the point on BF, and the quadrilateral aefc is a diamond, eh ⊥ AC, and the perpendicular foot is h EH = 1 / 2cf,

It is proved that in square ABCD, AC ⊥ BD, AC = BD, OB = 1 / 2, BD = 1 / 2, AC = BD = 1 / 2, ab = 1 / 2, ab = 1 / 2,
The quadrilateral aefc is rhombic,
∴AC=CF,AC∥EF,
∵EH⊥AC,
∴∠BOH=∠OHE=∠OBE=90°,
The quadrilateral beho is a rectangle,
∴EH=OB,
∴EH=1/2 AC=1/2 CF．

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

Two diagonal lines AC and BD of parallelogram ABCD intersect at O, ab = radical 5, Ao = 2, OB = 1

Suppose AC ⊥ BD
Then △ AOB is a right triangle
So OA ^ 2 + ob ^ 2 = AB ^ 2
That is 4 + 1 = 5
So AB = √ 5, which is the same as the given condition in the stem
So the hypothesis holds
That is, AOB = 90 degree
AC ⊥ BD
I don't know what knowledge points you have learned. Can you understand them?

In square ABCD, point F is on BC, CE is perpendicular to DF, intersects AB with point E, if CE = 10, find the length of DF

Let CE and DF intersect at point o,
Because CE is vertical to DF, it is easy to prove that triangular COF is similar to triangular CDF
Because CE is vertical to DF, it is easy to prove that triangular COF is similar to triangular ECB
So triangle CDF is similar to triangle ECB
Because it is a square, the triangle CDF and the triangle ECB have two corresponding sides, CB = CD
So the triangle CDF is congruent with the triangle ECB, so DF = CE = 10