As shown in the figure, the two diagonals of rectangle ABCD intersect at point O, through O for of ⊥ ad at point F, of = 2, through a for AE ⊥ BD at point E, and be: BD = 1:4, find the length of AC

As shown in the figure, the two diagonals of rectangle ABCD intersect at point O, through O for of ⊥ ad at point F, of = 2, through a for AE ⊥ BD at point E, and be: BD = 1:4, find the length of AC

∵ the quadrilateral ABCD is a rectangle, ∵ OA = ob, ∵ be: BD = 1:4, ∵ let be = x, then BD = 4x, ∵ OE = 4x-2x-x = x, ∵ be = OE, and ∵ AE ⊥ BD, ∵ ABO is an equilateral triangle, ∵ OA = AB, ∵ of ⊥ ad, of = 2, ∵ of is the median line of △ abd, ∵ AB = 2of = 2 × 2 = 4, ∵ AC = 2oa = 2Ab = 2 × 4 = 8

Two diagonals of rectangle ABCD intersect at O, of is perpendicular to ad, F, of = 2, AE is perpendicular to BD, e, be: BD = 1:4, find the length of AC

Obviously AB = 4
Let be = x, then BD = 4x
Because AB ^ 2 = be * BD
So 16 = 4x ^ 2
x=2
So AC = BD = 4x = 8

It is known that the diagonal lines AC and BD of rectangular ABCD intersect at O, of is perpendicular to AD and F, AE is perpendicular to BD and E, and be: ED is equal to 1:3
It is known that: the diagonal lines AC and BD of rectangular ABCD intersect at point O, of is perpendicular to AD and F, AE is perpendicular to BD and E, and be: ED is equal to 1:3. If AC is equal to 2 cm, of is equal to (), AE is equal to ()

∵AE⊥BD,∠BAD=90°
∴AE²=BE*ED
∵BE：ED=1:3
∴AE=√3BE
∴∠BAE=30°
∴∠OAF=30°
∵AC=2
∴AO=1
Of = 1 / 2oa = 1 / 2cm, AE = √ 3 / 2cm
Thank you, tiger bat

It is known that in rectangular ABCD, AE is perpendicular to BD and E, diagonal acbd intersects o, and be is equal to 1:3 AD and 6cm longer than ed

∵ rectangle ABCD
∴AO＝BO＝BD/2
∵BE：ED＝1：3
∴ED＝3BE
∴BD＝BE+ED＝4BE
∴BO＝BD/2＝2BE
E is the midpoint of Bo
∵AE⊥BD
The AE vertical bisection Bo
∴AB＝BO
Ψ equilateral △ abo
∴∠ABD＝60
∴AB＝AD/√3＝6/√3＝2√3
∴AE＝AB×√3/2＝2√3×√3/2＝3（cm）
The math group answered your question,

Two diagonals of rectangle ABCD intersect at O, AE vertical BD, of vertical ad, be: ed = 1:3, of = 2 to find AC

Let be = x, then ed = 3x;
Triangle AOF meets triangle ACD;
Because rectangle ABCD is divided into O points and diagonal lines;
Because of = 2, CD = 4
Because triangle AEB is similar to triangle BCD (right angles are equal AB / / DC, angle abd = angle BDC)
So x over 4 = 4 over 4x
X=2
Then AC = 8

In rectangular ABCD, AE is perpendicular to e, diagonal AC, BD intersects at point O, be: ed = 1:3, ad = 6cm
A little trouble!

In a right triangle bad, because AE is perpendicular to e,
So ad * ad = de * BD;
According to be: ed = 1:3, let be = x, then ed = 3x, BD = 4x;
So 6 * 6 = (3x) * (4x)
The solution is x = radical 3, that is, be = radical 3, de = triple radical 3;
And AE * AE = be * De, then
AE * AE = 9, so AE = 3

It is known that the diagonals of rectangle ABCD intersect at O, of vertical ad, of = 2cm, AE vertical BD, and be: ed = 1:3
The answer is 4cm,

AB=2OF=4cm;
Let AC = BD = x;
Then 4 ^ 2 - (x / 4) ^ 2 = (x / 2) ^ 2 - (x / 2-x / 4) ^ 2
We get x = 8

The diagonals AC and BD of rectangle ABCD intersect at O, AE ⊥ BD at e, of ⊥ ad at F. given be: ed = 1:3, of = 4cm, find the length of AC

16cm
OB = OD, be: ed = 1:3
BE=EO
Again AE ⊥ BD
So AB = Ao
From OA = ob
OA=OB=AB
So AOF = 30 degrees
AC=16cm

As shown in the figure, in rectangle ABCD, if OE ⊥ AC crosses e through diagonal o, then the length of AE is
Steps to take

No graph, no number, but the solution is as follows:
⊙ o is the midpoint of AC, OE ⊥ AC
The OE is the vertical bisector of AC
∴AE=CE
Let AE = CE = X
Then we can use Pythagorean theorem to set up equations in right triangle CDE

In trapezoidal ABCD, AB flat DC, angle a = 60 degrees, angle = 45 degrees, DC = 2, ad = 4, calculate the area of trapezoidal ABCD

Six times three plus six