As shown in the figure, AB / / CD, ad vertical DC, ab = BC, and AE vertical BC 1. Prove that ad = AE If ad = 8, DC = 4, the length of AB is 4

As shown in the figure, AB / / CD, ad vertical DC, ab = BC, and AE vertical BC 1. Prove that ad = AE If ad = 8, DC = 4, the length of AB is 4

Make the extension line of BF ⊥ DC to F
∵ ABCD is a right angle trapezoid, AB / / CD, ad ⊥ DC
The ABFD is a rectangle,
∴ AD=BF,
∴∠BCF=∠ABE
∵∠AEB=∠BFC=90°,
AB=BC
∴ △AEB≌△BFC
∴ AE=BF
∴AE=AD
2、
Let AB = BC = X,
Make CK perpendicular to AB and AB to point K
Then there is: in the triangle BCK, there is Pythagorean theorem, 4 ^ 2 + (x-0.5) ^ 2 = x ^ 2, x = 16.25
AB=16.25

In the parallelogram ABCD, AE is perpendicular to BC to e, AF is perpendicular to CD to F, EAF is equal to 45 degrees, and AE plus AF is equal to 2 times the root sign Find the circumference of ABCD

In the parallelogram ABCD, AE is perpendicular to BC to e, AF is perpendicular to CD to F, EAF is equal to 45 degrees, and AE + AF = 2 times root 2, then what is the circumference of the parallelogram ABCD?
∵∠EAF=45°
AE⊥BC
AF⊥CD
Ψ C = 135 ° (the sum of inner angles of quadrilateral is 360 °)
Ψ B = ∠ d = 45 °, C = ∠ bad = 135 ° (parallelogram)
∠ BAE = ∠ fad = 45 ° (the sum of internal angles of triangles is 180 °)
ν AB = AE, AF = DF (isosceles triangle)__
∵∠B=∠D=45°,AE+AF=2√2 __
∴AB+AD=4 (sin45°=√2／2)
The circumference is 4 × 2 = 8

As shown in the figure, in the isosceles right triangle ABC, ad ⊥ BC, PE ⊥ AB, PF ⊥ AC, then ⊥ DEF is______ Triangle

Δ DEF is an isosceles right triangle,
The reasons are as follows.
∵ △ ABC is an isosceles right triangle, ad ⊥ BC,
∴∠DAF=∠B=45°AD=BD，
∵PE⊥AB，PF⊥AC，
The aepf is a rectangle,
∴AF=PE，
∵∠B=45°，
∴PF=BE，
∴AF=BE，
In △ bed and △ AFD,
BD＝AD
∠B＝∠DAF
BE＝AF ，
∴△BED≌△AFD（SAS），
∴DE=DF，∠BDE=∠ADF，
∵∠BDE+∠ADE=90°，
∴∠ADF+∠ADE=90°，
∴∠EDF=90°，
The △ DEF is an isosceles right triangle,
So the answer is: isosceles right angle

As shown in the figure, in △ ABC, ab = AC, point P is any point on line BC (different from points B and C), PE ∥ AC intersects AB with E, PF ∥ AB intersects AC with F. what is the quantitative relationship between segments PE, PF and ab? Tell me your reasons

A: pf + ab
It is proved that in △ ABC, ab = AC,
∴∠B=∠C，
∵PE∥AC，PF∥AB，
The quadrilateral aepf is a parallelogram, ∠ BPE = ∠ C,
∴AE=PF，∠B=∠BPE，
∴BE=PE，
∴PE+PF=AE+BE=AB．

As shown in the figure, it is known that ab = AC = 2 ∠ BAC = 90 ° in RT △ ABC, P is a moving point on the hypotenuse BC, PE ⊥ AB, PF ⊥ AC, connected EF, D are the midpoint of BC edge As shown in the figure, we know that in RT △ ABC, ab = AC = 2, ∠ BAC = 90 °, P is a moving point on the hypotenuse BC, PE ⊥ ab, PF ⊥ AC, connected with EF, D is the midpoint of BC edge, (1) Find the length of hypotenuse BC (2) Judge the quantitative and positional relationship between de and DF, and explain your reasons (3) Find the area of quadrilateral AEDF As shown in the figure, we know that in RT △ ABC, ab = AC = 2 ∠ BAC = 90 ° P is a moving point on the hypotenuse BC, PE ⊥ ab, PF ⊥ AC, connected with EF, D is the midpoint of BC edge, (1) Find the length of the hypotenuse BC. (2) Judge the quantitative and positional relationship between de and DF, and explain your reasons. (3) Find the area of the quadrilateral AEDF.

1. ∵ AB = AC = 2 ∵ BAC = 90 ° ᙽ BC ∵ ab ∵ AC ∵ 2 ᙽ BC = 2 √ 22, connecting ad ? D is the midpoint of the BC edge, △ ABC is an isosceles right triangle ? ad = 1 / 2BC = BD  CAD = ﹤ ABC = 45 ° i.e. ? PE ⊥ AB, PF ⊥ AC  pea = ∠ AFP = ∠ ba

As shown in the figure, in △ ABC, the angle c = 90 ° AC = BC, D is the midpoint of AB, P is any point on AB, PE ⊥ AC, PF ⊥ BC, perpendicular foot e, F What is the size relationship between the segments de and DF? Does this relationship change with the movement of point P, and explain the reasons

De = DF, and the relationship remains unchanged
Reason: according to the meaning of the title, the quadrilateral epfc is rectangular, so PE = CF
Because ∠ C = 90 degrees, AC = BC
So ∠ a = ∠ B = 45 degrees
Because PE ⊥ AC, △ AEP is an isosceles right triangle, so AE = PE = CF
Because D is the midpoint of hypotenuse ab
So ad = DC = dB
So ∠ DCB = ∠ B = 45 degrees
In △ AED and △ CDF
AE = CF, ∠ DCB = ∠ a = 45 degrees, ad = DC
So △ AED ≌ △ CDF
So de = DF
In this question, the triangle DEF is always an isosceles right triangle. If you are interested, you can prove it yourself

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 3, BC = 4, the point P is any point on the AB side, passing through P is PE ⊥ AC in E, PF ⊥ BC in F, then the minimum value of line EF is______ .

Connect CP, ∵ ACB = 90 °, AC = 3, BC = 4. According to Pythagorean theorem, ab = 5, ∵ PE ⊥ AC, PF ⊥ BC,

As shown in the figure, in △ ABC, ab = 6, AC = 8, BC = 10, P is a moving point on edge BC, PE ⊥ AB is at e, PF ⊥ AC is at F, M is the midpoint of EF, then the minimum value of AM is () A. 2 B. 2.4 C. 2.6 D. 3

Connect AP, in △ ABC, ab = 6, AC = 8, BC = 10,  BAC = 90 °, ∵ PE ⊥ AB, PF ⊥ AC, ᚉ quadrilateral afpe is a rectangle, ᙽ EF = AP.

As shown in the figure, △ ABC, ad is its angular bisector, P is a point on ad, PE ∥ AB intersects BC and E, PF ∥ AC intersects BC and F. verification: the distance from D to PE is equal to that from D to PF

Proof: ∵ PE ∥ AB, PF ∥ AC,
∴∠EPD=∠BAD，∠DPF=∠CAD，
In ABC, ad is its angular bisector,
∴∠BAD=∠CAD，
∴∠EPD=∠DPF，
That is, DP bisection ∠ EPF,
The distance from D to PE is equal to that from D to PF

As shown in the figure, in △ ABC, ab = 3, AC = 4, BC = 5, P is a moving point on edge BC, PE ⊥ AB is in E, PF ⊥ AC is in F, then the minimum value of EF is () A. 2 B. 2.2 C. 2.4 D. 2.5

In △ ABC, ab = 3, AC = 4, BC = 5,
∴AB2+AC2=BC2，
In other words, ∠ BAC = 90 °
⊥ f ⊥ f ⊥ P ⊥ f ⊥ P ⊥ P ⊥ f ⊥ P ⊥ f ⊥ P ⊥ f ⊥ f,
The aepf is a rectangle,
∴EF=AP．
Because the minimum value of AP is the height on the hypotenuse of the right triangle ABC, which is 2.4,
The minimum value of EF is 2.4,
Therefore, C