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Mathematical problems of geometry in the second day of junior high school
In triangle ABC, AB=AC. D is a point on the extension line of CB. The angle ADB=60 degrees, e is a point on AD, and there is DE=DB. Verify AE=BE+BC.

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