In the trapezoidal ABCD, AD BC, A 90, 45, E is the midpoint of CD, AB 2AD 4, find the length of BE

After the point d, the left df is perpendicular to bc and the perpendicular foot is the point f, because ad bc, a=90° ab=2ad=4, so df=4 bf=2 bc=6, and because c=45° so fc=4, so from the pythagorean theorem, dc=4 2 e is the midpoint of cd, then ec=2 2, so in bec and the cosine theorem can be be 2=ec 2+bc 2-2ec*bc*cos c, since bc ec c is known, so it can be solved be=2 5 I hope I wish you success in your studies.

In trapezoidal ABCD, AD BC, A=90°, C=45°, E is the midpoint of Cd, AB=2AD=4, extend CD and BA to intersect to point F, cross point E to make the line segment EG parallel to BC, because C=45°, the triangle BCF is an isosceles right triangle, AD BC, BC=BF, and then according to the similarity between the small triangle and the large triangle, the value of each line segment can be obtained, AB=4, AD=2, AF=2, BF=AB+AF=6, BC=bf=6。 According to the trilateral relationship of the right triangle, we can deduce be=2 5 (2 times the root number 5).I can't send you the whole process in a way, but this problem is very good after doing the auxiliary line segments, I hope it can help you.

This problem is solved by using the characteristics of triangles and right triangles, you can try to see if you can make it. (Be sure to make the auxiliary line segment I mentioned).

Solution: If the lead accompanies the extension line of CD to BA in F, it is easy to know: Qiaoshu ADF and BCF are isosceles right triangles, which can be known:

df= 2, ad bc=df cf= 2 3 2=1 3, ad=1, then bc=3, isosceles right-angled acacia wide stupid triangle bce, can be found: be=3 2 2

Three-thirds of the root number two, write it as it reads. After a little d to make dg bc in g, it came out by tinkering with it yourself.

From the inscription, we know that the triangle BEC is an isosceles right triangle, BC=3, then BE=3 is 2 under the root

Solution: The parallel line of DC through B intersects the extension line of DA at M, and the extension line of DM is intercepted by M=CE

Then the quadrilateral mdcb is a square, and we can get mnb and ceb, so b e=bn and nbe=90°

abe=45°

abe= abn, then nab eab set ec=mn=y,ad=a

Then am=a,de=2a-y,ae=an=a+y, ad 2+de 2=ae 2,a 2+(2a-y) 2=(a+y) 2, and y=2a 3

Thus mn=2a 3,bm=dc=2*ad=2atan aeb=tan bnm=bm mn=2a (2a 3)=3

If you pass point A to do af bc to f, and to sail with ag be at g point a so that ad=a, then ab= 5 a

Let ec=x, according to the cosine formula, we get a 2+(2a-x) 2=5a 2+4a 2+x 2-2 5 a (4a 2+x 2)*cos45°

Solvable x=2a3

ag=bg=√10 a/2 be=2√10 a/3ge=be-bg=√10 /6

tan aeb=ag state Li stupid wheel ge=3

Extend a, b, and c, e at point f.

E is the midpoint of AD.

DE=EAABCD is trapezoidal and DC parallel AB

d=∠eaf

In AEF and DCE.

fea = dec (equal to the vertex angle).

ea=deeaf=∠d（asa）

AEF is fully equal to DCE

fe=ec，fa=dc

and dc=fa=1, ab=2, ab+fa=cb=3 (isosceles triangle) and fe=ec

be CE (3-in-1).

Take the midpoint F of BC and connect EF, and the source of the cavity is EF is the median line of the trapezoid, EF=(ab+cd) 2

and ab+cd=1+2=3=bc, ef=bc2, bec is rt, and bfc=90°, be ce

If you have any questions, please ask; If you are satisfied, thank you!

Proof: pass point C as CF AD and AB to F

And because ab cd so the quadrilateral adcf is a parallelogram, then ad=cf and af=cd=1, and because ab 3, if bf=1 and because a=90°, cf ab because bc=3, so in the triangle cfb, ad 2=cf 2=3 2-1 2, so ad=cf=2 is root two.

And because e is the midpoint of AD, de=root-two, in the triangle dec, CE 2=2+1=3 so CE=root-three. In the same way, we get be = root 6

And because BC 2 3 2 = CE 2 + BE 2, the triangle is a right triangle and CE be

Take C as AB's high argument or liter CF A=90° AB CD, D=90° ADCF as the moment to carry the old form.

bc=3 bf=ab-cd=1

CF = change sign 8 ae = change sign 2

BE = 6 CE = 3 Silver

ce^2+be^2=bc^2

ce⊥be

Solution: Pass the point d to make de ab to hand over bc to e

Ad BC, de Nalu AB

Parallelogram abed

ad＝5,ab＝4

Be ad 5, de hu osmotic ab 3

Again de ab

dec＝∠b

b+∠c＝90

dec+∠c＝90

cde＝90

CD 4CE (DE +CD ) (9+16) 5BC BE+CE 5+5 10

Answer: bc=10

Analysis: It has long been referred to as d as a state shirt de ab to get a parallelogram abed so Lu closed with de = ab = 3 , be = ad = 5

Because: b+ c=90°

So: dec+ c=90°

So: get the right triangle cde

Again: de = 3 , cd = 4; So: ce = 5

So: bc = be + ec = 5 + 5 = 10

You're not right, isn't that right? The inner angles of the quadrilateral are 360 degrees, b+ c=90°, then the sum of the remaining two angles is 270 degrees. It can also be said that the width difference is mine, and my knowledge is not enough to help the landlord.