Mathematics proof questions for the first year of junior high school, mathematics proof questions fo

Bisected. The proof is as follows: eg, ad are both perpendicular to BC so the two sides are parallel, and according to the theorem of parallel lines, the angle e is equal to the angle 3

Angle 1 is equal to angle 2

And because the angle e is equal to the angle 1

So angle 2 is equal to angle 3

Gu Dezheng.

Because ad bc is in d, eg bc is in g

So 3= ceg 1= 2

And because e= 1 1= 2

So 3= ceg 1= 2

Because 3= 2 then ad divides the bac

Although it has been proven, it is not possible to divide this question equally in the spirit of responsibility, and the reason why it can be proved should be because of the additional condition of e= 1, so in line with the principle of responsibility, this question is a wrong question.

Solution: Because ad bc is in d, eg bc is in g

So ad parallel eg

So 1= 2, e= 3

again e= 1

So 2= 3

i.e. AD bisects the BAC

Because ad bc is in d, eg bc is in g

So ad eg (reason omitted).

So angle 3 = angle e

And because 1= e

2+∠3=∠1+∠e

So 2= 3

So AD divides the bac equally

Deuce. E is equal to 1 and its opposite vertex angles to give e and b are congruent, and e is congruent with c, so c = b, and because ad bc, we get 2 = 3

Because eg ad;

Therefore<3=and because <1=, <2=<3;

So divide it equally ...

Because ad bc and eg bc

So the EG is parallel to the AD

1=∠2，∠e=∠3

Because e= 1

So 2= 3

So AD divides the bac equally

Both EG and AD are perpendicular to BC, so EG and AD are parallel, angle E and angle 3 are equal, angle 1 and angle 2 are equal, so angle 2 and angle 3 are equal.

For the center of the circle, the appropriate length is the radius of the arc intersection ab, ac in e, as the line segment ab = as the center of the circle ae as the radius arc, intersection ab in the center, ef is the semi-longitude arc and the front arc intersect in the connection af, take the point c on the af, so that ac = even bc, the triangle abc is sought. Evidence: ab=ab; angle a=a; ac=ac, so the triangle abc is equal to abc(sas).

60°。dcf= dea (obtained in parallel), in the same way eab= cfb, fcb=60°, b=90°, then eab=30°. and ae bisect dab, then dae=30°.

d=90°。then dcf=60°

1.AC is the bisector of the angle dab, where angle 1 = angle 2 angle 2 = angle 4, angle 1 = angle 4, ab is parallel to dc 2ab dc, ad bc, abcd are parallelograms angle 1 = angle 3, angle 2 = angle 4

Angle 1 = angle 2

Angle 3 = Angle 4

The minimum perimeter of the triangle PQR is.

10 is dotted p

About the symmetry points P1 and P2 of OA and OB

Easy to prove.

Triangle PQR circumference = P1Q+QR+RQ2

Due to the two points in between.

The straight line is the shortest. So only.

When p1, q, r, p2

When in a straight line, p1q+qr+rq2

The shortest connection OP1 and OP2 is required

The angle p1oq2 = 60°

The easily obtained triangle p1oq2 is an equilateral triangle.

According to the triangular congruence theory.

op=op1

And because the triangle p1oq2 is an equilateral triangle.

So p1p2=op=op1=10

So the minimum circumference of the triangle PQR is 10

Crossing the point E to do EF parallel to AB

then bef= abe

bef+∠def=∠bed

bed= abe+ cde so def= cde so cd parallel ef

So ab parallel cd

Solution: AB is parallel to CD

Cross the point e to make parallel lines of ab and cd.

abe+∠cde =∠beg+∠deg∠bed=∠beg+∠deg

ab//cd

Haha, revisit the junior high school topic

My method goes like this, after e as a cd parallel line ef, then the parallel line property is known def= cde, because the title says bed= abe+ cde

So abe= bef, so ab ef cd

Feel free to ask

60 degrees. b=90, bcf=60, so cfb=, so eab= cfb=bisect bad, so dae= eab=30d=90, so dea=60, ae cf, so dcf= dea=60

Solution: Because bcf=60°, b=90°

So afc= bcf+ b=150°

Because of AE CF

So cfb= eab=30°

Because AE divides the bad

So bae= dae=30°

Because d=90°

So aed=60°

Because, bcf=60°,ae cf,so, eab= cfb=30°

Again, ae bisects bad, so, dae=30°, so, dea=60°=dcf

So, dcf = 60°

f and c are both redundant with c ef, so f= c, and b= edf, bc=df so abc edf, get ac=ef

As shown in the figure, it is known that the triangle ABC and the triangle def are a puzzle of a pair of triangles, a, e, c, and d are on the same straight line. 1) Verify EF parallel BC; 2) Find the degrees of angle 1 and angle 2.

From the meaning of the title: a= b=45°, d=60°, f=30°】1) Prove that the triangle ABC and the triangle def are a puzzle of triangles, fec= ecb=90°, fec+ ecb=180°, ef bc

2) Solution: 2= d+ qcd=60°+90°=150° epb= a+ aep=45°+90°=135° In the pentagonal epoqc, 1=540°- pec- ecq- 2- epo=540°-90°-90°-150°-135°=75°

What version of the book did you use, and what geometry did you learn?

Let's take the textbook seriously, it's all the essence.

If you knew this time, why bother in the first place?

2）：∠apm=∠apn

Suppose that the line segment AC and the line segment PN intersect at the point O, then there is (because AC=AN, the angle NAB=angle CAM (because NAB=60+BAC, MAC=60+BAC), ab=AM) the triangle MAC congruence and the triangle BAN, so there is the angle MCA=angle BNA, and because the angle AOC=POC, there is a triangle AON, which is similar to the triangle POC

AO=ON oc, and because the angle AOP=ANGLE NOC, the triangle APO is similar to the triangle NOC, so APN=ACN=60

Similarly, apm=60

∵ ac||de

acd= d, acb= e (1) and acd= b

d=∠b （2）

and ac=ce (3).

According to the corners of (1), (2), (3).

Get: ABC CDE

I don't know if you understand o( oha!

Proof: Because of AC DE

So d= acd= b and acb= e

Because ac=ce

So abc cde (corner corner).

Proof: Because of AC De, in ABC and CDE, ACB = DEC

And because acb= dec, acd= b, ac=ce

So abc cde

Because ac de, so, acb= e,, acd= d. Because, acd= b, so b= d. And because a+ b+ acb=180, d+ e+ dce=180, a= dce.

To sum up, ac=ce, acb= e, a= dce, so abc cde(asa).

Because AC is the bisector of the angle DAB.

So angle 2 = angle 1

Because angle 2 = angle 4

So angle 1 = angle 4

So AB is parallel to DC.

Because ab dc, ad bc

So angle 1 = angle 4 angle 2 = angle 3

Because angle 1 = angle 2

So angle 3 = angle 4

1) 2 = 4, ac is the bisector of the dab (known) 1 = 2 (definition of the bisector of the angle).

1 = 4 (equal substitution).

ab dc (equal inner wrong angles, two straight lines parallel).

2) 1 = 2, 2 = 4 (known).

1 = 4 (equal substitution).

ab dc (known).

2 = 3 (two straight lines are parallel, and the inner wrong angles are equal).

again 2= 4, 2= 3 (known).

3 = 4 (equal substitution).

1.AC is the bisector of the angle dab, where angle 1 = angle 2 angle 2 = angle 4, angle 1 = angle 4, ab is parallel to dc.

2.ab dc, ad bc, abcd are parallelograms.

Angle 1 = Angle 3, Angle 2 = Angle 4

Angle 1 = angle 2

Angle 3 = Angle 4

AB is parallel to DC.

Reason: .AC is the bisector of the angle dab, where angle 1 = angle 2

Angle 2 = Angle 4, Angle 1 = Angle 4, AB is parallel to DC.

Proof : AC df, a ADF 180° (two parallel lines, complementary to the side inner angles) and a 1, 1 ADF 180°, cf ae, (complementary to the side side of the inside angles, two lines parallel) and 3 4, cb ef quadrilateral CFEB is a parallelogram, e 2

Because AC parallel DF

Angle A = Angle FDE

Because angle 1 = angle a

So angle 1 = angle fde

So CF is parallel to DE

Because angle 3 = angle 4

So CB is parallel to EF

So CBEF is a parallelogram.

So angle e = angle 2

∠a=∠1,∴cf‖ad

3=∠ 4,∴bc‖ef

cf‖ad,bc‖ef

The quadrilateral BCFE is a parallelogram.

e=∠2