Grade 9 Mathematics Geometry Proof Questions, Middle School Math Geometry Proof Questions

Because ab=ac=bc

So, angular BAC = angular ACD = 60 degrees.

Because ae=cd

So the triangle ABE is exactly equal to the triangle CAD (corner edge), so the angle AEB = the angle CDA

Because the angle APE = 180-angle AEB-angle EAP, the angle ACD = 180-angle CDA-angle EAP

So, the angle ape = angle acd = 60 degrees.

So, the angle bpq = 60 degrees.

Because of bq ad

So, the angle pbq = 30 degrees.

So bp=2pq

Proof: Because abc is an equilateral triangle.

Easy Abe ADC

So adb= bap+ pae=60°

abe=30°，bqa=90°，so

bp=2pq (the edge of the 30° angle is half of the hypotenuse).

The idea is that because ab=ac=bc, abc is an equilateral triangle with three angles of 60°

and ae=cd so abe= adc so

It should be a positive angle bpq = 60 degrees or an angle pbq = 30 degrees.

The intersection point of de and ac is g

Connecting the AD angle DAB is equal to the angle DAF plus the angle 45 degrees.

The angle def plus 45 degrees equals the angle CGD equals the angle ABC

Equal to the angle DEA plus 45 degrees.

Outer angles, inner angles, and theorem.

The midline of the hypotenuse in a right triangle is equal to half of the hypotenuse.

AD is equal to 1/2

BC so the angle DAB is equal to the angle DBA

So the angle DAE is equal to the angle DEA

So AD is equal to DE is equal to 1/2BC

From a and d perpendicular lines to bc, through the triangle congruence theorem, it can be known that abc and bcd are equal, according to the known conditions be=cf, bc edges are common, then abc and bcd are equal, using the triangle congruence theorem (corner edges), bec= cfb can be proved.

Solution: Use the corner edge to verify that the triangle BEC is similar to the triangle CBF because: ab=dc ad bc

So: trapezoidal ABCD is isosceles trapezoidal.

So: angular cbe = angular bcf

Because :be=2ea cf=2fd ab=dc so:be=fc

Because: BC=CBoCBc=AngleBCF=CBF=CF, the triangle BEC is similar to the triangle CBF

According to the similarity triangle theorem.

So the angle bec = the angle cfb

Proof: AD BC, ab DC, trapezoidal ABCD is isosceles trapezoidal, ABC= DCB.

be=2ea,cf=2fd，ab＝dc，∴eb=2/3ab,fc=2/3dc，∴eb=fc.

eb=fc，bc=cb，∠abc=∠dcb，∴△ebc≌△fcb，∴∠bec=∠cfb。

As can be seen from the title, the trapezoidal ABCD is an isosceles trapezoid.

and BE 2EA, CF 2FD

So there's be cf

So in BEC vs. CFB.

bc＝bcbe＝cf

ebc＝∠fcb

So the triangle BEC is all equal to the triangle CFB

So bec cfb

Prove that the triangle BEC is equal to the triangle CFB, and the two sides and their angles are equal!!

ok！！！

Proof: Because of ad||bc,ab=dc.So abc= bcd because the points e and f are on ab and dc respectively, and be=2ea, cf=2fd ab dc

So be=cf bc edges are shared, so the triangle bce is all equal to the triangle bcf

So bec cfb

Because ad bc, ab dc so abc= dcb, because ab dc, be=2ea, cf=2fd so be=cf, and because bc=bc so ebc is all equal to fbc, so bec= cfb

Proof: Because ab=dc, be=2ea, cf=2fd, be=cf

In the triangle bec and triangle cfb, bc is the common side, and the trapezoidal abcd is an isosceles trapezoid, abc= dcb, be=cf, so the triangle bec is all equal to the triangle cfb, so bec= cfb

ad bc,ab dc so abc= dcbab dc,be=2ea,cf=2fd so be=cfbc=bc so ebc is all equal to fbc

bec=∠cfb

Do it one question at a time. Pretend to be a stool block 1

de//bc

The triangular coarse form ADE is similar to the triangular ABC.

Let de be x and make the triangle ade high h

The scale factor is k

Hall Hu. BC is KX

The height of the triangle BEC is HK-H, i.e. H(K-1).

then Syntagonium hx=8

hxk（k-1）=48

Solve it yourself.

Solution: Make a perpendicular to BC over A

Then bf = 3 so af = change sign 3

ab = 2 times change sign 3

Because dc=ad

And because of the triangle ACD triangle BCA

So the square of AC = CD CB

So cd=2=ad

That's the final answer.

Take advantage of the trigonometric functions of the third year of junior high school. First pass the fixed point A to do AD perpendicular to BC and hang the foot for point E. I link AD and do it by a characteristic of the right triangle of 30 degrees.

In the case of an isosceles triangle, the foot is 30°. The length of the high ae is calculated according to the trigonometric function. It's root number 3

It's so simple that you can get a triangle and ade is also a 30° right triangle. Utilize a 60° angle sin

One time. The answer can be calculated!

The answer is: 2

If C is done CE perpendicular to QP and D then the triangle CDQ and triangle CEQ congruence, and the triangle CBP and triangle CEP congruence. So qp=qe+ep=dq+bp

Triangle apq perimeter = dq + aq + ap + ab = 22Extend the CF, cross M to make JK, cross the CF extension line to J, cross BE to K, and JK is perpendicular to CF and BE.

Angle BMK = Angle CMJ, BM=CM, Angle MBK = Angle MCJ So triangle BMK congruent triangle CMJ

mj=mk, the quadrilateral jkef is rectangular, so jf=ek, angle mjf=angle mke=90 degrees.

So the triangle mjf congruent triangle mke

So me=mf

The answer is: 150°.

Let the side length of the square be 1, because cde is an equilateral triangle, so de=dc=ad=1. So ade is an isosceles triangle.

Because 1=30°, dae= dea=75°, so eab=15°, and because eab is an isosceles triangle, so 3=150°

tan=opposite than the edge of the forest?

cot=hypotenuse than forest edge?

sin=opposite than hypotenuse?

cos=forest edge than hypotenuse?

I forgot about it every month after the exam.

Because ABCD is a square, AB=BC=CD=AD and because the triangle CDE is an equilateral triangle, EC=CD=DE, so the angle 1=angle 2=90 degrees-60 degrees=30 degrees AD=DE, EC=BC That is, the triangle ADE and the triangle BCE are equilateral triangles, so the angle DAE=Angle DEA=Angle CEB=Angle CBE=75 degrees, and the angle EAB= Angle ABE=90 degrees-75 degrees=15 degrees.

In the triangle ABE, angle 3 = 150 degrees.

You see, the triangle CDE is an equilateral triangle, then the angle EDC = angle ECD = 60 degrees, then the angle 1 = angle 2 = 30 degrees, and because ed=dc=ad=bc, the triangle ADE and the triangle BEC are congruent and isosceles triangles, i.e., the angle DAE = angle CBE=75 degrees, so the angle EAB = angle EBA = 15 degrees, then angle 3 is naturally equal to 150 degrees.

150 degrees Remember to give points Because the triangle CDE is an equilateral triangle and the quadrilateral ABCD is a square, angle 1 is 30 degrees. ad=cd=de, so the angle AED is 75 degrees, and the same angle BEC is 75 degrees, so angle 3 is 150 degrees (the inner angle of the quadrilateral and 360 degrees, and the inner angle of the equilateral triangle is 60 degrees).

Because cde is an equilateral triangle, de equals ce, equals cd, so dce is 60 degrees, 3 is 30 degrees, because quadrilateral abcd is square, so cb equals cd equals ce, so ceb equals cbe equals 75 degrees, because abc is equal to 90 degrees, so abe is equal to 15 degrees (the other is the same), so 3 is equal to 150 degrees. It's all clear enough in words, right!!

Actually, it's very simple, and I'm not going to prove it to you.

Because abcd is square, ad=dc

Yes Because DCE is an equilateral triangle ed=dc angle edc=60 so angle 1=90 minus 60=30°

Because ad=dc ed=dc so ad=de, so the angle aed=angleead=(180-30) divided by 2=75, so the angle3=360-60-75-75=150 don't understand q404485684

Solution: Because the quadrilateral ABCD is a square and the triangle CDE is an equilateral triangle, then ade bce(sas),ad=ed=dc=ec=bc, 1= 2=90°, so ae be, so eab= eba, so dae= cbe=75°

So eab= eba=15°

So 3= eab- eba=1350°

1) Proof: ABCD is square.

ad=bc，∠adc=∠bcd=90°

Another triangle CDE is an equilateral triangle.

ce=cd，∠edc=∠ecd=60°

ade=∠ecb

ade≌△bce．

2) Solution: CDE is an equilateral triangle, CE=CD=BC CBE is an isosceles triangle, and the vertex angle ECB=90°-60°=30° EBC= 1 2(180°-30°)=75°AD BC AFB= EBC=75°

Analysis: (1) From the side AD=BC of the subject square ABCD, in the equilateral triangle CDE, CE=DE, EDC is equal to ECD, that is, its congruence can be proved

2) According to the angular relationship between equilateral triangles, isosceles triangles, and parallel lines, the degree of AFB can be obtained

Isn't there a lack of conditions?

Yes. The triangular EOB is similar to the triangular ODA.

Proof: ac=ab

a=∠b∠c=90°

b=∠a=45°

aoe=∠b+∠oeb=∠doa+∠doe∠b=∠doe=45°

oeb=∠doa

a=∠b ∠doa=∠oeb

Triangle EOB Triangle ODA

2) Triangle EOB Triangle ODA

be/ob=ao/ad

ac=bc=2 ∠c=90°

ab=8½ao=bo=2½

be=y ad=x

yx=ao*bo=2

y=2／x(1＜x＜2)

3) Isosceles triangle be=ad when x=y=2 when triangle ode

b=∠aao=bo

Triangle ebo triangle DAO

EO=DO triangle ODE is an isosceles triangle.

Actually, it's very easy, the O point is the side of the ECD, beo, oed, and aod must be similar,

Extension of AF to A, AFB congruence A Fe, A E=Ab=Da, AE=Ca, A Ea = 180- Bae= DaC, A Ea Congruent DAC, Aa E= CDA, BAA = CDA

Angular baa + fad = 90, cda + fad = 90

Extend af to a triangle afb congruent triangle a fe, a e=ab=da, ae=ca, a ea=180- bae= dac, a ea congruent dac, aa e= cda, baa = cda

Angular baa + fad = 90, cda + fad = 90