Junior 2 Math and Geometry Proof Questions?

Satan

25. In ABCD, the points E and F are respectively in AB, and on CD DF=BE, is the quadrilateral DEBF a parallelogram? Tell me your reasoning

Test Points: Determination and properties of parallelograms.

Topic: General question type.

Analysis: From the meaning of the question, a set of quadrilaterals with parallel and equal opposite sides can be judged to be parallelograms Answer: Solution: Yes

Reason: In the parallelogram ABCD, then AB CD, and AB=CD, and DF=BE, the quadrilateral debf is a parallelogram

Comments: This question mainly examines the judgment of parallelograms, and should be proficient

About what? Equation? Geometry? Function? Vector?

It should be classified, right? Easy or hard?

Two right-angled triangles have two corresponding equal heights and one equal angle. Ask them to be all-in.

You've learned!!

Because: in the parallelogram ABCD.

So; ao=co

Because: ace is an equilateral triangle.

So: aeo = CEO (3 in 1).

Because: ae=ce

Because: ed=de

So: ade= cde

So: ad=cd

So: the parallelogram ABCD is a diamond.

2) Because: the triangle ace is an equal triangle.

So: aed= ced=30°

Because: aed=2 ead

So: ead=15°

Because: the parallelogram ABCD is a diamond.

So: bao= dao

Because: eac=60°

So: dao=45°

So: bad=90°

So: the diamond-shaped ABCD is a square.

I'm also in my second year of junior high school, so it's a bit hard to think about it. Just give a little hard work.

1. Proof that because the triangle ace is an equilateral triangle, ae=ce, and because abcd is a parallelogram, ao=co, ad=bc, ab=dc and eo=e'o, so by de=de', so ad=cd, i.e., ad=bc=ab=dc

Therefore the quadrilateral ABCD is rhomboid;

2. Proof: Because the triangle AEC is an equilateral triangle, and the proved EO in proof 1 is the midline of the triangle AEC, so aed=30°, ead=15°, adb= aed+ ead=45°, and because proof 1 knows that the quadrilateral ABCD is a diamond, the quadrilateral ABCD is a square.

The triangle abe cbe congruence be ae=ce ao=oc indicates that be bisected angle e

ab=bc So the four sides are equal The parallelogram is a diamond, sorry I just copied it and missed it.

Question 2: Ace equilateral triangle According to question 1, we know that the angle aed ced is 30 degrees, and then the angle ead ecd is 15 degrees, so the angle ado cdo is 45 degrees, adding up to 90, so the diamond is a square.

Proof: 1The parallelogram ABCD introduces the triangle AOB which is all equal to the triangle COD (the symbol can't be played, don't mean it).

Then we deduce AO=OC, and then according to the triangle AEC is an equilateral triangle, EO is the midline of the triangle AEC.

According to the inference of the unity of the three lines of the equilateral triangle, AEO= OEC=30 degrees, and AE=EC, plus BE=BE, the triangle AEB is all equal to the triangle CEB, and then AB=BC, and the quadrilateral is a diamond.

2. 1.If AED=2 EAD, then EAD=15, AD=135, according to AE=EC, AED= CED, DE=DE, we get that the triangle AED is all equal to the triangle CED, ADE= CDE=135 degrees, and ADC=180- ADE= CDE=90 degrees. Finally, it is concluded that the quadrilateral abcd is a square.

Proof: Because p is any point on the angular bisector of abc, then pb=pc (the distance from the point on the angular bisector to both sides of the angle is equal) because ab ac

ab-ac＞pb-pc

de ac then fdc= ecd; and cd bisects acb then the angle fcd= ecd, so fdc= fcd, i.e., fc=fd;

In the same way, fc de, we can get de=ec; In the triangle DFC and DEC, dc is shared, FCD = ECD; fdc = edc, then the two triangles are congruent, in summary, fc=ce=de=df; If all four sides are equal and the opposite sides are parallel, then the quadrilateral decf is a diamond.

Known: In the right triangle ABC, ACB=90°. m is the midpoint of ab, d is the point of the extension of bc, and cd=bm, and b= d

Hope you understand.

Let the two bisected angles be x and y.

Proof: Both abc and cde are equilateral triangles.

bc=ac，cd=ce，∠acb=∠dce=60°∴∠bcd=∠ace

bcd≌ace

cae=∠b=60°

cae=∠acb

ae‖bc

Proof that ABC and EDC are equilateral triangles, acb= dce=60°, ac=bc, dc=ec and bcd= acb- acd, ace= dce- acd, bcd= ace

ace≌△bcd．

ace bcd, abc = cae = 60°, and acb = 60°, cae = acb, ae bc

Because the angle BAC = angle DEC, the triangle ADO (tentatively O) is similar to CEO, the angle DCB = angle ace, extending BA, CE to F, the angle DAE is equal to the angle FAC angle FAE is equal to 60 degrees is equal to the angle B, so BC AE

It's diamond-shaped because they overlap each other when folded.

Then aef is equal to def

So ae=de

AF = DF Angle AEF = Angle DEF

Angular AFE = Angular DFE

Angular EAF = Angular EDF

ab = 2, bc = 2 root number 3, ac = 4

then it turns out that ABC is a right triangle.

AB BC because of FD BC

then ab fd

So the angle AEF = the angle DFE

Angular EAF = Angular cab

Because angular AFE = angular DFE

Angular EAF = Angular EDF

then the angle AEF = the angle AFE

Angle cab = angle edf

So AEF is an isosceles triangle.

af ed, so ae=af

The quadrilateral aedf is a parallelogram.

Because ae=de

af=df, so ae=de=af=df

So the quadrilateral aedf is a rhombus.

Let the middle point be o

Ado Eco

then ao eo=do co

ao/do=eo/co

aoe=∠doe

aoe∽△doc

eao=∠cdo=60°=∠acb

ae∥bc