answer
Because: angle EDB = 60 DE = decibel.
So: △EDB is an equilateral triangle, and DE=DB=EB.
A vertical line that crosses a when BC crosses BC at F.
Because: △ABC is an isosceles triangle
So: BF=CF, 2BF=BC.
Also: the angle DAF=30 degrees.
So: AD=2DF.
Also: DF=DB+BF
So: AD=2(DB+BF)=2DB+2BF=2DB+BC.
(AE+ED)=2DB+BC, where ED=DB.
So: AE=DB+BC, AE=BE+BC. It is known that the AB side and AC side of △ABC are taken as sides, and squares ABED and ACFG are made outward respectively, and points P, Q, O 1 and O2 are the midpoints of DG, BC, DB and GC respectively. Verification: Quadrilateral O 1QO2P is the square answer connecting DC and Q02p.
Then, using neutral line, we get PO 1 parallel GB parallel O2Q, PO 1 parallel and equal to O2Q, PO2 parallel DC parallel O 1Q, PO2 parallel and equal to O1q.
Because DC is vertical and equal to BG, the quadrilateral O 1QO2P is a square.
Trapezoidal ABCD
Parallel BC
AD=AB=DC
BD vertical CD
If the trapezoid perimeter is 10.
Verify the degree of angle c.
Trapezoidal area
Rectangular ABCD
AB=5
BC= 12
Alternating current
Bachelor of science
Hand over to o
P is a little higher than BC.
PM vertical BD
PN vertical AC
Find the value of PM+PN
Answer 1)cos
C=-cosA
AD=X
BD=X*tanC
BD^2=X^2+X^2-2*X*X*-cos
C
x^2*(tanc)^2=x^2+x^2-2*x*x*-cos
C
1-(cosC)^2=(2+2cosC
)(cosC
)^2
(cosc+ 1)[2(cosc)^2+cosc- 1]=0
cosC≦- 1，cosC & gt0
So:) [2 (cosc) 2+cosc- 1] = 0.
cosC= 1/2，cosC=- 1
C=60
BC=2DC
DC= 10/5=2, BC =4.
Height H=√3
S trapezoid =(BC+AD)*H/2
=3√3
2)
BD= 13
PM/DC=BP/BD，PN/AB=PC/AC，AC=BD，AB=CD
(PM+PN)/AB=BC/BD
PM+PN=60/ 13 It is known that in BE⊥AD parallelogram ABCD, point M is the midpoint of DC and AB=2AD.
Verification: ∠EMC=3∠DEM The answer is MO⊥BE, connect BM.
BE⊥AD,MO⊥BE, so: Mo is parallel to AD and BC, ∠ DEM = ∠ ome, ∠ BMO = ∠ CBM.
M is the midpoint of CD, so O is the midpoint of BE. It is deduced that BMO of two triangles is equal to EMO, so: ∠ ome = ∠ BMO.
AB=2AD
M is the midpoint of DC, so: BC = cm, then the triangle CBM is isosceles. So ∠ CBM = ∠ BMC.
So: ∠ DEM = ∠ ome = ∠ BMO = ∠ BMC.
And: ∠ ome+∠ BMO+∠ BMC = ∠ EMC.
So: ∠EMC=3∠DEM