As shown in the figure, in the triangle ABC, ab = BC, angle ABC = 90, there is a straight line PQ passing through point B, ad is made through a and C respectively, PQ is vertical to D, CE is vertical to E (1) (2) if the line PQ intersects with the line AC, other conditions remain unchanged, whether the conclusion in (1) is still valid, if not, write the corresponding conclusion and prove it

As shown in the figure, in the triangle ABC, ab = BC, angle ABC = 90, there is a straight line PQ passing through point B, ad is made through a and C respectively, PQ is vertical to D, CE is vertical to E (1) (2) if the line PQ intersects with the line AC, other conditions remain unchanged, whether the conclusion in (1) is still valid, if not, write the corresponding conclusion and prove it

(1)DE=AD+CE
First, it is proved that △ abd is equal to △ BCE (right triangles have one side equal and one diagonal complementary)
Then ad = be
CE=BD
Add the two equations
(2)
If PQ and AC intersect
Then De is equal to the absolute value of the difference between AD and CE
Same as (1), it is proved that △ abd is equal to △ BCE
AD=BE
CE=BD
Suppose ad > CE
Just subtract the two equations

In △ ABC, ad is high, vertex P and N of rectangular pqmn are on AB and AC respectively, and QM is on edge BC. If BC = 8cm, ad = 6cm, and PN = 2pq, the perimeter of rectangular pqmn is calculated

According to the meaning of the question, if PQ: ad = BP: AB, PN: BC = AP: AB, pqad + & nbsp; pnbc = bpab + apab = AP + pbab = ABAB = 1, and ∵ PN = 2pq, BC = 8cm, ad = 6cm, ∵ pq6 + 2pq8 = 1, ∵ PQ = 2.4, then PN = 4.8, ∵ perimeter of rectangular pqmn = 14.4cm

In the acute triangle ABC, the bottom edge BC = 12, BC height BD = 8, intercept rectangle pqmn, QM on BC, PN on AB, BC respectively. How to make the rectangle area maximum
I hope the prawns can be more detailed. Benzene and I can't understand if we don't understand~
Please try to use the knowledge of grade three! This is not a quadratic function@@

Let QM intersect ad with E, AE = x, Mn = ed = 8-x
Because QM is parallel to BC, AE: ad = QM: BC
That is, X: 8 = QM: 12
QM=3X/2
Rectangular area y = 3x / 2 * (8-x) = - 3x square / 2 + 12x
The maximum value is the ordinate of the vertex of the quadratic function = 24

As shown in the figure, the right angle vertex B of isosceles RT △ ABC is on the line PQ, ad ⊥ PQ is on D, CE ⊥ PQ is on e, ab = BC. Prove de = AD + CE

It's a proof of complete equality~
Certificate:
∠DBA+∠CBE=∠ABC=90°
∵AD⊥PQ CE⊥PQ
∴∠DAB+∠DBA=∠CBE+∠BEC=90°
∴∠DAB=∠CBE
Because ∠ DAB = ∠ CBE
AB=AC
∠DBA=∠BCE（ASA）
∴△ADB≌△BCE

As shown in the figure, △ ABC BC = 60cm, height ad = 40cm, quadrilateral pqmn is a rectangle, point P is on AB side, points Q and m are on BC side, and point n is on AC side. (1) if PQ: PN = 1:3, find the length of each side of the rectangle. (2) let PN = x, PQ = y, find the functional relationship between Y and X

(1) Let PQ = k, PN = 3k, ∵ quadrilateral pqmn be a rectangle, ∵ PN ∥ BC, ∵ aead = pnbc, ∵ BC = 60cm, ad = 40cm, ∵ 40 − K40 = 3k60, the solution is k = 403, 3K = 3 × 403 = 40. ∵ the lengths of each side of the rectangle are 403cm, 40cm, 403cm, 40cm; (2) ∵ quadrilateral pqmn be a rectangle, ∵ PN ∥ BC, ∵ aead = pnbc, ∵ PN = x, PQ = y, ∵ 40 − Y40 = X60, and y = - 23x + 40 The formula is y = - 23x + 40

As shown in the figure ad is the high point of the triangle ABC, PQ is on the BC side, M is on the AC side, BC = 30, ad = 20, and the quadrilateral pqmn is a square
Find the side length of pqmn

Let the side length of the square be x and the intersection of AD and Mn be e
⊙ square pqmn, ad ⊥ BC
Ψ rectangular pned
∴ED＝PN＝X
∴AE＝AD-ED＝20-X
∵MN∥BC
∴MN/BC＝AE/AD
∴X/30＝（20-X）/20
X＝12
A: the side length of a square pqmn is 12
The math group answered your question,

M as shown in the figure, in the isosceles triangle ABC, the bottom edge BC = 60cm, the height ad = 40cm, and the quadrilateral PQRS is a square. (1) is △ ARS similar to △ ABC?
(2) Find the side length of a square pqsr
As shown in the figure, in △ ABC, de ‖ BC, s △ ADE:S Trapezoid bced = 1:4, find ad: DB

As shown in the figure, let rs = x, & nbsp; & nbsp; AE / ad = RS / AB & nbsp; & nbsp; & nbsp; (40-x) / 40 = x / 60 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 24, ② AD & # 178;: ab & # 178; = 1 ∶ 5 & nbsp; & nbsp; & nbsp; & nbsp

In isosceles trapezoid ABCD, ad ∥ BC, ab = DC, diagonal AC, BD at O, and ∠ AOB = 60 °, e, F, G are the midpoint of OA, ob, CD respectively

If angle AOB is equal to 60, then the triangle AOB is an equilateral triangle and F is the midpoint of Ao, so BF is perpendicular to Ao, (three lines in one) in right triangle BFC, G is the midpoint of BC, so FG is equal to half of BC (right triangle three lines in one)

In isosceles trapezoid ABCD, DC / / AB and ab > DC, ad = BC, diagonal AC and BD intersect at O, angle AOB = 60 degrees, m, N and P are the midpoint of OD, OA and BC respectively
It is proved that △ MNP is an equilateral triangle

It is proved that the base angles of BN and cm are equal in the isosceles trapezoid ABCD, so the angle OAB = angle oba, the angle doc = angle COD and the angle AOB = 60 degrees. According to the angle cod = 60 degrees, the triangle AOB and the triangle COD are regular triangles. Because n is the middle point of Ao, BN is the middle line of the side Ao of the triangle AOB, so BN is the triangle AOB

As shown in the figure, in the equilateral trapezoid ABCD, AD / / BC, ad = 3, BC = 5, AC and BD intersect at point O, and ∠ BOC = 60 °. If AB = CD = x, then the value range of X is

Calculate BD = 8, so AB > bd-ad, and ab is less than 8, 5 < x < 8